Solved Name Genetics Pedigree Worksheet A pedigree is a | Chegg.com - Free Printable

Let's solve the Genetics Pedigree Worksheet step by step, based on the image you described.

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Part 1: Analyze the Pedigree

We are given a pedigree chart with symbols:

- ○ = female, unaffected
- ● = female, affected
- □ = male, unaffected
- ■ = male, affected

Each generation is labeled with Roman numerals (I, II, III), and individuals are numbered within each generation.

#### Pedigree Summary:

- Generation I:
- I-1: Male, affected (■)
- I-2: Female, unaffected (○)

- Generation II:
- II-1: Male, unaffected (□)
- II-2: Male, unaffected (□)
- II-3: Female, unaffected (○)
- II-4: Male, unaffected (□)
- II-5: Female, unaffected (○)
- II-6: Male, unaffected (□)
- II-7: Female, unaffected (○)
- II-8: Male, unaffected (□)
- II-9: Male, unaffected (□)

- Generation III:
- III-1: Male, unaffected (□)
- III-2: Male, unaffected (□)
- III-3: Female, affected (●)
- III-4: Male, affected (■)
- III-5: Female, unaffected (○)
- III-6: Male, unaffected (□)
- III-7: Female, unaffected (○)
- III-8: Male, unaffected (□)

Note: The only affected individuals in this pedigree are:
- III-3 (female)
- III-4 (male)

Their parents are:
- II-3 (female, unaffected) and II-4 (male, unaffected)

So, two unaffected parents have two affected children.

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Question: Try to identify the genotypes of the following individuals using the pedigree above.

We assume the trait is genetic, and we need to determine whether it's dominant or recessive first.

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1. Is this trait dominant or recessive? Explain your answer.

✔ Answer: Recessive

Explanation:
The trait appears in III-3 and III-4, who are both affected, but their parents (II-3 and II-4) are unaffected. This means the trait must be recessive, because if it were dominant, at least one parent would have to show the trait (since a dominant allele would express itself).

In autosomal recessive inheritance, two carriers (heterozygous) can produce an affected child (homozygous recessive). This fits perfectly here.

> Rule: If unaffected parents have an affected child → the trait is recessive.

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2. How can you know for sure that individuals II-3 and II-4 are heterozygous?

✔ Answer:
Because they are unaffected (so they do not have the disease), yet they had two affected children (III-3 and III-4). For a recessive trait, both parents must carry at least one copy of the recessive allele to pass it on.

Since both children are homozygous recessive (aa), they must have inherited one recessive allele from each parent. Therefore, both II-3 and II-4 must be heterozygous (Aa).

If either parent were homozygous dominant (AA), they couldn’t pass on the recessive allele, so no affected children would result.

Thus, II-3 and II-4 must be heterozygous (Aa).

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Now, determine genotypes:

Let’s define:
- A = dominant, normal (unaffected)
- a = recessive, affected

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#### III-3:
- Affected → aa
✔ Genotype: Homozygous recessive

#### II-1:
- Unaffected, but his parents are I-1 (affected, ■) and I-2 (unaffected, ○)
- I-1 is affected → genotype = aa
- I-2 is unaffected → could be AA or Aa
- But since I-1 is aa, all children must get a from him
- II-1 is unaffected → must be Aa (because he got 'a' from father, and 'A' from mother to be unaffected)

✔ Genotype: Heterozygous

#### I-1:
- Affected → aa
✔ Genotype: Homozygous recessive

#### II-4:
- We already established: unaffected, but has affected children → must be Aa
✔ Genotype: Heterozygous

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✔ Final Answers:

| Individual | Genotype Type |
|----------|----------------------------|
| III-3 | Homozygous recessive |
| II-1 | Heterozygous |
| I-1 | Homozygous recessive |
| II-4 | Heterozygous |

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3. Brown eyes are dominant (B), blue eyes are recessive (b).

Scenario:
- A brown-eyed woman whose father had blue eyes.
- She marries a brown-eyed man whose parents were also brown-eyed.
- They have a blue-eyed son.

We are to draw a pedigree showing:
- All four grandparents
- The two parents
- The son
- Indicate genotypes where certain, or possible

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Step-by-step analysis:

#### Blue-eyed son → bb
- So, both parents must have contributed a b allele.

#### Mother:
- Brown-eyed → B_ (could be BB or Bb)
- Her father had blue eyes → bb
- So, she must have inherited b from her father
- Since she is brown-eyed, she must have B from her mother
- So, her genotype is Bb

✔ Mother: Bb

#### Father:
- Brown-eyed → B_
- His parents were both brown-eyed → both B_
- But he has a blue-eyed son (bb) → so he must be Bb (to pass b)
- Could his parents be BB or Bb?

Let’s consider:
- Both parents brown-eyed → could be BB or Bb
- But since their child (the father) is Bb, then at least one parent must be Bb.
- But we don't know which — so both possibilities exist.

So, father’s genotype: Bb

But his parents: both brown-eyed → possible genotypes: BB or Bb

But since they produced a Bb child, at least one must be Bb.

So, possible combinations:
- One parent: BB, other: Bb → child could be Bb
- Or both: Bb → child could be Bb

So, we cannot be certain of the grandparents’ genotypes.

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Draw the Pedigree:

```
Grandparents (Generation I):

[I-1] [I-2]
(♂, B_) (♀, B_)
| |
| |
| |
[II-1] [II-2]
(♂, Bb) (♀, Bb)
| |
| |
| |
[III-1]
(♂, bb)
```

Wait — let's clarify:

Actually, the mother is the daughter of a blue-eyed father.

So:

#### Mother's side:

- Grandfather: blue-eyed → bb
- Grandmother: brown-eyed → B_ (but must be Bb, since she passed B to daughter)
- Daughter (mother): brown-eyed → Bb (must be, because she got b from father)

#### Father's side:

- Grandfather: brown-eyed → B_ (unknown: BB or Bb)
- Grandmother: brown-eyed → B_ (unknown: BB or Bb)
- Son (father): brown-eyed → Bb (because he passed b to son)

So, the father’s parents must both be at least one Bb, but we don’t know which.

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Final Pedigree:

```
Generation I (Grandparents):

[I-1] [I-2] [I-3] [I-4]
(♂, bb) (♀, Bb) (♂, ?) (♀, ?)
| | | |
| | | |
| | | |
[II-1] [II-2] [II-3] [II-4]
(♀, Bb) (♂, Bb) (♂, ?) (♀, ?)
| | | |
| | | |
| | | |
[III-1]
(♂, bb)
```

Wait — better labeling.

Let’s define:

- Mother = II-1 (daughter of blue-eyed father)
- Father = II-2 (son of brown-eyed parents)
- Child = III-1 (blue-eyed)

So:

```
Generation I:

[I-1] [I-2] [I-3] [I-4]
(♂, bb) (♀, B?) (♂, B?) (♀, B?)
| | | |
| | | |
| | | |
[II-1] [II-2] [II-3] [II-4]
(♀, Bb) (♂, Bb) (♂, ?) (♀, ?)
| | | |
| | | |
| | | |
[III-1]
(♂, bb)
```

But actually, II-1 and II-2 are the parents, and III-1 is their son.

So:

Corrected Pedigree:

```
Generation I:

[I-1] [I-2] [I-3] [I-4]
(♂, bb) (♀, B_) (♂, B_) (♀, B_)
| | | |
| | | |
| | | |
[II-1] [II-2] [II-3] [II-4]
(♀, Bb) (♂, Bb) (♂, ?) (♀, ?)
| | | |
| | | |
| | | |
[III-1]
(♂, bb)
```

Now label genotypes:

- I-1 (mother's father): blue-eyed → bb
- I-2 (mother's mother): brown-eyed → must be Bb (since daughter is Bb and got b from father; she gave B to daughter)
- So I-2: Bb
- I-3 (father's father): brown-eyed → B_ → could be BB or Bb → unknown
- I-4 (father's mother): brown-eyed → B_ → could be BB or Bb → unknown
- II-1 (mother): brown-eyed, daughter of bb × Bb → genotype: Bb
- II-2 (father): brown-eyed, but has blue-eyed son → must be Bb
- III-1 (son): blue-eyed → bb

So, certain genotypes:

- I-1: bb
- I-2: Bb
- II-1: Bb
- II-2: Bb
- III-1: bb

Uncertain genotypes:

- I-3: B_ → could be BB or Bb
- I-4: B_ → could be BB or Bb

But since they had a child (II-2) who is Bb, at least one of them must be Bb.

So possible combinations:
- I-3: BB, I-4: Bb → possible
- I-3: Bb, I-4: BB → possible
- I-3: Bb, I-4: Bb → possible

So no way to be certain of I-3 and I-4's genotypes.

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Final Answer for Question 3:

Pedigree:

```
Generation I:
[I-1] (♂, bb) [I-2] (♀, Bb) [I-3] (♂, B?) [I-4] (♀, B?)
\_____________/ \_______________/
\ /
\ /
[II-1] (♀, Bb) [II-2] (♂, Bb)
\__________________/
|
|
[III-1] (♂, bb)
```

Genotypes:
- Certain:
- I-1: bb
- I-2: Bb
- II-1: Bb
- II-2: Bb
- III-1: bb

- Uncertain (possible genotypes):
- I-3: BB or Bb
- I-4: BB or Bb

(At least one of I-3 or I-4 must be Bb.)

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✔ Summary of Answers:

#### Part 1: Genotype Identification

| Individual | Genotype Type |
|-----------|--------------------------|
| III-3 | Homozygous recessive |
| II-1 | Heterozygous |
| I-1 | Homozygous recessive |
| II-4 | Heterozygous |

#### Question 1: Dominant or Recessive?
- Recessive — because unaffected parents (II-3 and II-4) have affected children (III-3 and III-4).

#### Question 2: Why are II-3 and II-4 heterozygous?
- Because they are unaffected but have affected children. For a recessive trait, both parents must be carriers (heterozygous) to produce an affected child.

#### Question 3: Eye Color Pedigree
- See diagram above.
- Certain genotypes: I-1 (bb), I-2 (Bb), II-1 (Bb), II-2 (Bb), III-1 (bb)
- Uncertain: I-3 and I-4 → both could be BB or Bb (but at least one must be Bb)

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