Pedigree analysis exercise for genetics education.

Let’s go step by step through each question on the pedigree analysis worksheet.

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Question 1:
> How many individuals are in Generation I? ___
> How many in Generation II? ___
> How many in Generation III? ___

Looking at the first pedigree (top left):

- Generation I: Top row — has 2 people (one square, one circle) → 2
- Generation II: Middle row — has 4 people (two squares, two circles) → 4
- Generation III: Bottom row — has 3 people (one square, two circles) → 3

✔ Answers:
Gen I: 2
Gen II: 4
Gen III: 3

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Question 2:
> There are no carriers for Huntington’s Disease because it is a _______________ disorder.

Huntington’s disease is caused by a dominant allele. That means if you have even one copy of the bad gene, you get the disease. So there’s no such thing as a “carrier” who doesn’t show symptoms — everyone with the gene shows it.

✔ Answer: dominant

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Question 3:
> Who in this family has Huntington’s Disease? ________________________

In the first pedigree, shaded symbols = affected.

- Gen I: The male (square) is shaded → he has it.
- Gen II: Two females (circles) are shaded → they have it.
- Gen III: One female (circle) is shaded → she has it.

So list them by generation and position:

→ I-1 (male), II-2 and II-3 (females), III-2 (female)

But since the blank probably expects names or labels like “I-1, II-2, etc.” — we’ll use standard notation.

✔ Answer: I-1, II-2, II-3, III-2

*(Note: Sometimes worksheets expect just numbers or positions — but based on typical format, using Roman numeral + dash + number is correct.)*

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Question 4:
> Is Huntington’s disease dominant or recessive? _______________

We already said it’s dominant — because affected parents can pass it to kids, and unaffected people don’t carry it silently.

Also, in the pedigree, every affected person has an affected parent (except maybe I-1, who could be the original mutation). But more importantly, unaffected people never have affected children unless the other parent is affected — which fits dominant inheritance.

✔ Answer: dominant

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Question 5:
> If individual #8 marries someone who does not have Huntington’s, what are the chances their child will have Huntington’s?

First, find individual #8.

In the second pedigree (middle right), individuals are numbered 1–16.

Look at the chart:

- Individual #8 is in Generation III, third from left — it’s a shaded circle → so she HAS Huntington’s.

Since Huntington’s is dominant, let’s assign alleles:

Let H = Huntington’s allele (dominant)
h = normal allele (recessive)

Affected people must have at least one H.

Individual #8 is affected → genotype is either HH or Hh.

But look at her parents:

Her father is #4 (unaffected square → hh)
Her mother is #3 (affected circle → must be H_, but since her husband is hh, and some kids are unaffected, she must be Hh)

Wait — actually, let’s trace carefully.

In Pedigree 2:

- #3 (female, affected) married #4 (male, unaffected → hh)
- Their children: #7 (unaffected male → hh), #8 (affected female → must be Hh, because dad gave h, mom gave H), #9 (unaffected female → hh), #10 (affected male → Hh)

So yes — #8 got H from mom (#3) and h from dad (#4) → so #8 is Hh

Now, she marries someone without Huntington’s → that person is hh

Cross: Hh x hh

Offspring possibilities:

- 50% Hh → affected
- 50% hh → unaffected

✔ Answer: 50% or 1/2

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Question 6:
> Why do all the daughters in generation II carry the colorblind gene?

This refers to the bottom pedigree (colorblindness).

Colorblindness is X-linked recessive.

Symbols:

- Square = male
- Circle = female
- Shaded = affected
- Half-shaded or dot inside = carrier (for X-linked traits)

In the bottom pedigree:

Generation I:

- Male #1: shaded → colorblind → genotype X^c Y
- Female #2: not shaded, but has a dot → carrier → X^C X^c

Their children (Gen II):

All daughters get:

- From dad: X^c (since he only has X^c to give to daughters)
- From mom: either X^C or X^c

So all daughters are either X^C X^c (carrier) or X^c X^c (affected)

But in the diagram, all Gen II daughters are shown with dots → meaning carriers → so they must be X^C X^c

Why? Because mom gave them X^C, dad gave X^c → so all daughters are carriers.

Even though mom is carrier, she passed X^C to all daughters? Not necessarily — but in this case, the diagram shows all daughters as carriers (dots), not affected (fully shaded). So likely, mom passed X^C to all daughters.

Actually, looking again: In Gen II, there are 4 daughters — all have dots → all are carriers.

That means each daughter received:

- X^c from dad (he’s colorblind → only X^c)
- X^C from mom (so mom must have given X^C to all 4 daughters — possible by chance, or perhaps the diagram implies that)

But the reason ALL daughters carry the gene is because:

→ Dad is colorblind → gives X^c to all daughters
→ Mom is carrier → can give X^C or X^c
→ But in this family, all daughters ended up getting X^C from mom → so they’re all carriers (X^C X^c)

The key point: Since dad is colorblind, he passes his X chromosome (with mutant allele) to ALL his daughters. So every daughter gets at least one copy of the colorblind allele → so they all carry it (even if not affected).

✔ Answer: Because their father is colorblind and passes his X chromosome with the colorblind allele to all his daughters.

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Question 7:
> Why is individual #13 colorblind?

Individual #13 is in Generation IV, last row, far right — shaded square → male, colorblind.

His parents:

- Mother: #9 (in Gen III) — circle with dot → carrier → X^C X^c
- Father: #10 (in Gen III) — unshaded square → normal → X^C Y

Son #13 is male → gets Y from dad, X from mom.

Mom is carrier → can give X^C or X^c

He got X^c from mom → so he is X^c Y → colorblind.

✔ Answer: He inherited the colorblind allele (X^c) from his mother, who is a carrier.

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Question 8:
> Write the genotypes for individuals #1 and #2.

Again, bottom pedigree — colorblindness (X-linked recessive)

#1: male, shaded → colorblind → genotype: X^c Y

#2: female, not shaded but has dot → carrier → genotype: X^C X^c

✔ Answer:
#1: X^c Y
#2: X^C X^c

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Question 9:
> What are the genotypes of individuals #3 and #4?

#3: male, unshaded → normal vision → X^C Y

#4: female, unshaded, no dot → NOT carrier → must be X^C X^C

Wait — check their children.

They have:

- #5: male, shaded → colorblind → X^c Y → got X^c from mom → so mom MUST be carrier!

Contradiction?

Hold on — let's re-express.

In bottom pedigree:

Gen I:

- #1: male, shaded → X^c Y
- #2: female, dot → carrier → X^C X^c

Gen II:

Children of #1 and #2:

- #3: male, unshaded → X^C Y (got X^C from mom)
- #4: female, unshaded, no dot? Wait — in the diagram, #4 is a circle with NO dot → so not carrier? But then how did her son #5 become colorblind?

Ah — here’s the issue. Let me double-check the image description.

Actually, in the user’s image description, for the bottom pedigree:

“#4” is listed as a circle — but in the text, it says:

> "individuals #3 and #4" — and #3 is male, #4 is female.

And their child #5 is shaded male → colorblind.

For #5 to be colorblind (X^c Y), he must have gotten X^c from his mother (#4).

Therefore, #4 MUST be a carrier → even if the diagram doesn’t show a dot, perhaps it’s implied, or maybe I misread.

Wait — looking back at the original problem statement in the image transcription:

It says:

> “Write the genotypes for individuals #1 and #2.” → done

> “What are the genotypes of individuals #3 and #4?”

And in the pedigree, #3 is male (unshaded) → X^C Y

#4 is female — and she has a colorblind son (#5), so she must be carrier → X^C X^c

Even if the symbol doesn’t have a dot, logically she must be.

Perhaps in this diagram, carriers aren't always marked with dots? Or maybe #4 is supposed to have a dot.

Given that #5 is colorblind, and father #3 is normal (X^C Y), then mother #4 must have given X^c → so she is carrier.

✔ So:

#3: X^C Y
#4: X^C X^c

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Question 10:
> Was individual #1’s mother a carrier of the colorblind gene? Explain.

Individual #1 is the colorblind male in Gen I → X^c Y

He got his X chromosome from his mother.

Since he is colorblind, his X chromosome carries the mutant allele → X^c

Therefore, his mother must have given him X^c → so she had at least one X^c allele.

She could be:

- Carrier: X^C X^c
- Affected: X^c X^c (but rare, and usually not shown unless specified)

But since the question asks if she was a *carrier*, and we know she passed X^c to her son, yes — she must have been at least a carrier.

Even if she were affected, she still carried the gene — but typically “carrier” implies heterozygous and not affected.

In X-linked recessive, males are affected if they have the allele; females need two copies to be affected.

So if #1’s mother was not affected (which we assume, since not stated otherwise), then she must have been a carrier.

✔ Answer: Yes, because individual #1 is colorblind (X^c Y), so he inherited the X^c allele from his mother. Therefore, his mother must have been a carrier (X^C X^c) or affected (X^c X^c). Since affected females are rare and not indicated, she was most likely a carrier.

But to keep it simple for student:

✔ Final simplified answer: Yes, because he got the colorblind gene from his mother, so she must have carried it.

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Now, compiling all answers clearly.

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Final Answer:

1. Generation I: 2, Generation II: 4, Generation III: 3
2. dominant
3. I-1, II-2, II-3, III-2
4. dominant
5. 50% (or 1/2)
6. Because their father is colorblind and passes his X chromosome with the colorblind allele to all his daughters.
7. He inherited the colorblind allele from his mother, who is a carrier.
8. #1: X^c Y, #2: X^C X^c
9. #3: X^C Y, #4: X^C X^c
10. Yes, because individual #1 is colorblind and must have inherited the colorblind allele from his mother, so she carried the gene.