Worksheet - Pedigree Practice Problems 2012 | PDF | Dominance ... - Free Printable

Let's solve this step-by-step based on the pedigree and the instructions provided.

---

Given:

- The trait is an autosomal recessive disorder.
- Squares = males, Circles = females.
- Shaded individuals = affected (have the disorder).
- Unshaded = unaffected (do not show the trait).

We are to:
1. Identify genotypes of all individuals.
2. Answer specific questions about generations, inheritance clues, hybrids, and genotype ambiguity.

---

## 🔍 Step 1: Understanding Autosomal Recessive Inheritance

For an autosomal recessive trait:
- Affected individuals must be homozygous recessive (aa).
- Unaffected individuals can be either:
- Homozygous dominant (AA) – not carriers
- Heterozygous (Aa) – carriers (but not affected)

---

## 🧬 Genotype Assignment Strategy:

We'll assign genotypes using:
- Affected → aa
- Unaffected but has affected children → must be Aa (carrier)
- Unaffected with no affected offspring and no known carrier parents → could be AA or Aa, but we’ll use logic to narrow down.

We will label each individual with their genotype.

---

## 🔢 Labeling the Pedigree Generations

Let’s number generations from top to bottom:

- Generation I: Individuals 1 & 2 (top row)
- Generation II: Children of I → 1, 2, 3, 4, 5
- Generation III: Children of II → 1–7 in third row
- Generation IV: Children of III → 1–3 in fourth row
- Generation V: Children of IV → 1 & 2 in fifth row

So, there are 5 generations.

---

## ✔ Part a: How many generations are presented?

> Answer: 5 generations

---

## ✔ Part b: Clues that the trait is recessive

We are told it's recessive, but let's confirm using clues in the pedigree:

Clues for autosomal recessive inheritance:

1. Affected individuals have unaffected parents
→ For example, Individual III-1 (black square) is affected (aa), but both parents (II-1 and II-2) are unaffected, meaning they must be carriers (Aa). This only happens if the trait is recessive.

2. Skipping generations
→ The trait appears in some generations but not others (e.g., Generation I has one affected, then skips to Generation III). This is typical of recessive traits.

3. Two unaffected parents produce an affected child
→ E.g., II-1 and II-2 are both normal, but have affected child III-1 → confirms both are carriers (Aa × Aa → 25% chance of aa).

4. Affected individuals are born to unaffected parents
→ Again, seen multiple times (e.g., III-4, IV-4, IV-5, V-1)

These are classic signs of autosomal recessive inheritance.

> ✔ Answer:
> The trait skips generations and appears in children of unaffected parents, indicating that the trait is recessive. Additionally, two unaffected parents can produce affected offspring, which only occurs when both parents are carriers (heterozygous).

---

## ✔ Part c: Shade half of circle/square for hybrid individuals

"Hybrid" means heterozygous (Aa) — carriers who do not show the trait.

We need to identify all heterozygous (Aa) individuals and shade half of their symbol (usually a half-shaded circle or square).

Let’s go through each individual and determine genotype.

---

🔹 Generation I

- I-1: Male, unaffected → must be Aa or AA
But his daughter (II-2) is unaffected, and her child (III-1) is affected → so II-2 must be Aa → therefore I-1 must be Aa (because if he were AA, he couldn't pass 'a' to make II-2 a carrier unless she got 'a' from I-2).
Wait — I-2 is affected, so she is aa. So she passed 'a' to all her children.

So I-1: unaffected, married to aa → his children:
- II-2: unaffected → must be Aa
- II-3: unaffected → must be Aa
- II-4: unaffected → must be Aa

Therefore, I-1 must be Aa (since he is unaffected, but had an affected wife and produced affected grandchildren via carrier daughters).

So I-1: Aa

- I-2: Female, affected → aa

✔ So I-1: Aa → shade half
I-2: aa → fully shaded

---

🔹 Generation II

Children of I-1 (Aa) and I-2 (aa):

Each child gets one allele from each parent.

- From I-2: always a
- From I-1: A or a → 50% chance

So all children are:
- If get A → Aa → unaffected (carrier)
- If get a → aa → affected

But in the chart:
- II-1: unaffected → Aa
- II-2: unaffected → Aa
- II-3: unaffected → Aa
- II-4: unaffected → Aa
- II-5: unaffected → Aa

Wait — all five are unaffected? But I-2 is aa, I-1 is Aa → expected: 50% Aa (unaffected), 50% aa (affected)

But in the pedigree, none of these five are shaded → all unaffected.

That means no aa among them → so all are Aa.

So:
- II-1: Aa
- II-2: Aa
- II-3: Aa
- II-4: Aa
- II-5: Aa

All are hybrids (Aa) → shade half of each.

---

🔹 Generation III

Now look at each couple.

#### Family 1: II-1 (Aa) × II-2 (Aa)

Children: III-1, III-2, III-3, III-4

- III-1: affected → aa
- III-2: unaffected → could be AA or Aa → but since parents are Aa × Aa, and III-1 is aa → both parents are Aa → III-2 must be Aa (can't be AA unless confirmed, but possible)
- However, III-2 has a child (IV-1) who is unaffected → but no info about whether IV-1 is carrier
- But III-2 is unaffected, and has affected sibling → must be Aa (because if she were AA, she couldn’t have an affected sibling unless both parents are Aa — which they are)
- So III-2: Aa
- III-3: unaffected → Aa or AA → but since both parents are Aa, and no other info, we can't rule out AA, but likely Aa → but we don't know for sure → so possibly ambiguous
- However, III-3 has no affected children → but his partner is unknown → so we can’t tell
- So possible Aa or AA
- III-4: affected → aa

So:
- III-1: aa
- III-2: Aa
- III-3: ? → could be AA or Aa → unknown
- III-4: aa

But wait — III-3 is unaffected, and his siblings include affected (III-1 and III-4) → so he must be Aa (cannot be AA, because if he were AA, he would not have received 'a' from parents, but parents are both Aa → so he could be AA or Aa)

But in autosomal recessive, unaffected individuals with affected siblings are almost certainly carriers (Aa) unless proven otherwise.

So we assume:
- III-3: Aa

→ So III-1: aa, III-2: Aa, III-3: Aa, III-4: aa

All except III-1 and III-4 are hybrids → shade half of III-2 and III-3

---

#### Family 2: II-3 (Aa) × II-4 (Aa)? Wait — II-3 and II-4 are siblings, but are they married?

No — II-3 is male, II-4 is male → can't marry → mistake.

Look again:

II-3: male, unaffected → Aa
II-4: male, unaffected → Aa
They are both sons of I-1 and I-2 → unrelated to each other

Then:
- II-3 is married to someone? No, no spouse shown → maybe he doesn't have kids?
- II-4 is married to II-5 (female, unaffected)

Yes! II-4 × II-5 → children: III-5, III-6, III-7

So:

#### Family: II-4 (Aa) × II-5 (Aa)

Children:
- III-5: unaffected → Aa or AA → but since both parents are Aa, and no affected children → could be AA or Aa → but we don’t know
- III-6: unaffected → same → Aa or AA
- III-7: unaffected → same

But none of them are affected → so no aa

So all three are unaffected, so genotypes: AA or Aa

But since both parents are Aa, each child has:
- 25% AA
- 50% Aa
- 25% aa

But none are affected → so all are either AA or Aa

But we cannot determine which → so genotype uncertain

So:
- III-5: A_ (AA or Aa)
- III-6: A_
- III-7: A_

But they could be carriers — so potentially Aa

But since we don’t know, we don’t shade them as hybrids unless we know.

But wait — later, III-8 is affected → and III-8 is son of II-4 and II-5 → so III-8 is aa

Therefore, both II-4 and II-5 must be Aa (since they are unaffected, but had affected child)

So:
- II-4: Aa
- II-5: Aa

Now, their children:
- III-5: unaffected → could be AA or Aa → but since parents are Aa × Aa → and child is unaffected → A_
- III-6: same → A_
- III-7: same → A_

But we don’t know if they are AA or Aa → so cannot confirm hybrid status

However, since they are unaffected and have affected siblings, they are likely carriers, but not certain.

But in standard pedigree analysis, unaffected individuals with affected siblings are considered carriers (Aa) unless evidence suggests otherwise.

So:
- III-5: Aa
- III-6: Aa
- III-7: Aa

→ So shade half of each.

But note: III-8 is affected → aa

So:
- III-8: aa

And III-9 is affected → aa

Wait — III-9 is female, affected → aa

But III-9 is daughter of II-4 and II-5 → same parents as III-5,6,7,8

So yes, III-9: aa

So now:

- III-5: Aa
- III-6: Aa
- III-7: Aa
- III-8: aa
- III-9: aa

So III-5,6,7 are hybrids → shade half

---

#### Family 3: III-8 (aa) × III-9 (aa)

Both affected → both aa

Children:
- IV-4: affected → aa
- IV-5: affected → aa

So both children are aa

No hybrids here.

---

#### Family 4: IV-5 (aa) × IV-6 (unaffected male)

IV-5: affected → aa

IV-6: unaffected → must be Aa or AA

But they have:
- V-1: affected → aa
- V-2: unaffected → Aa or AA

Since V-1 is aa, and IV-5 is aa, then IV-6 must have contributed a → so IV-6 must be Aa (since he is unaffected)

So:
- IV-6: Aa

Then:
- V-1: aa
- V-2: unaffected → must be Aa (since one parent is aa, other is Aa → child gets a from IV-5, and A or a from IV-6 → so if unaffected, must be Aa)

So:
- V-1: aa
- V-2: Aa

---

Now summarize all genotypes.

---

## ✔ Full Genotype List

Generation I:

- I-1: Aa (unaffected, carrier) → shade half
- I-2: aa (affected)

Generation II:

- II-1: Aa → shade half
- II-2: Aa → shade half
- II-3: Aa → shade half
- II-4: Aa → shade half
- II-5: Aa → shade half

Generation III:

- III-1: aa (affected)
- III-2: Aa → shade half
- III-3: Aa → shade half
- III-4: aa (affected)
- III-5: Aa → shade half
- III-6: Aa → shade half
- III-7: Aa → shade half
- III-8: aa (affected)
- III-9: aa (affected)

Generation IV:

- IV-1: unaffected → child of III-1 (aa) and III-2 (Aa) → must be Aa → shade half
- IV-2: unaffected → child of III-1 (aa) and III-2 (Aa) → must be Aa → shade half
- IV-3: unaffected → child of III-1 (aa) and III-2 (Aa) → must be Aa → shade half
- IV-4: aa (affected)
- IV-5: aa (affected)
- IV-6: Aa → shade half

Generation V:

- V-1: aa (affected)
- V-2: Aa → shade half

---

## ✔ Part c: Shade half for all hybrids (Aa)

So shade half of every individual who is Aa → listed above.

This includes:
- All unaffected individuals who are not AA → all except those with no risk

But in this case, every unaffected individual is Aa (except possibly some in generation III/IV where we might not know, but we deduced they are Aa due to family history)

So:
- Shade half of: I-1, II-1, II-2, II-3, II-4, II-5, III-2, III-3, III-5, III-6, III-7, IV-1, IV-2, IV-3, IV-6, V-2

(Do not shade III-1, III-4, III-8, III-9, IV-4, IV-5, V-1 — they are affected → already fully shaded)

---

## ✔ Part d: What is the genotype of the 6th individual in the fourth generation?

Fourth generation: IV-1, IV-2, IV-3, IV-4, IV-5, IV-6

So IV-6 is the 6th individual.

From earlier:
- IV-6: Aa (unaffected, married to IV-5 who is aa, and had affected child → must be carrier)

> ✔ Answer: Aa

---

## ✔ Part e: Explain why some individuals can have multiple genotypes. Provide an example.

In pedigrees, some individuals may have uncertain genotypes because:
- They are unaffected, so could be AA or Aa
- But we don’t know their exact genotype without more information (like offspring or testing)

However, in this case, most unaffected individuals are carriers because they have affected siblings or children.

But suppose an individual is unaffected and has no affected relatives and no affected children, we can’t be sure if they are AA or Aa.

But in this pedigree, all unaffected individuals have affected relatives, so we can usually infer they are Aa.

But let’s say III-5 — she is unaffected, but her parents are Aa × Aa, and she has an affected sibling (III-8). So she is very likely Aa, but could theoretically be AA — though unlikely.

But in practice, we assign Aa.

So the reason some individuals can have multiple genotypes is due to insufficient information.

> ✔ Example: Consider III-5. She is unaffected, but her parents are both carriers (Aa), and she has an affected sibling (III-8). While she is most likely Aa, it is possible she is AA (if she inherited A from both parents). Without knowing her offspring or DNA test, we cannot be certain — so her genotype is A_ (either AA or Aa).

Thus, some individuals can have multiple possible genotypes due to incomplete data.

> ✔ Answer: Some individuals can have multiple genotypes because their exact genetic makeup cannot be determined solely from the pedigree. For example, III-5 is unaffected but has an affected sibling, so she could be either homozygous dominant (AA) or heterozygous (Aa). Since we don’t know her offspring or perform genetic testing, we cannot definitively assign her genotype — thus, she has multiple possible genotypes.

---

## ✔ Final Answers

a. How many generations?

5 generations

b. Clues that trait is recessive?

The trait skips generations and appears in children of unaffected parents. Two unaffected parents can produce an affected child, which indicates both parents are carriers (heterozygous). This is characteristic of autosomal recessive inheritance.

c. Shade half of all hybrids (Aa)

Shade half of:
I-1, II-1, II-2, II-3, II-4, II-5, III-2, III-3, III-5, III-6, III-7, IV-1, IV-2, IV-3, IV-6, V-2

d. Genotype of 6th individual in 4th generation (IV-6):

Aa

e. Why some individuals have multiple genotypes?

Because their genotype cannot be definitively determined from the pedigree alone. For example, III-5 is unaffected but has an affected sibling; she could be AA or Aa. Without additional information (like offspring), her genotype remains uncertain.

---

Let me know if you'd like a visual version or a labeled pedigree!