pedigree practice answers - Name Date Period: Genetics Pedigree ... - Free Printable

Let’s solve this genetics worksheet step by step.

We’re looking at two pedigree charts and answering questions about how a trait is inherited — whether it’s dominant or recessive, and what genotypes people might have.

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Problem 1: First Pedigree (top one)

This chart shows a family with some members having a shaded symbol (affected) and others not. We’re told to assume the trait is rare unless stated otherwise.

Looking at generation I:
- Individual I-1 is affected (shaded square).
- Individual I-2 is unaffected (open circle).

They have children in generation II:
- Some are affected, some are not.

Now look at generation III:
- Two unaffected parents (II-5 and II-6) have an affected child (III-3). That’s key!

👉 If two unaffected parents can have an affected child, that means the trait must be recessive. Why? Because for a recessive trait, you need two copies of the bad gene to show the disease. Unaffected parents can still carry one copy each (carriers), and if both pass it on, the child gets sick.

So answer to part A: The trait is recessive.

Part B: What is the genotype of individual I-1?

Since the trait is recessive, and I-1 is affected, they must have two recessive alleles → let’s call them “a”. So genotype = aa

Part C: Can we determine the genotype of II-4?

II-4 is unaffected (open circle). But her father (I-1) is aa, so he must give her one “a” allele. Since she’s unaffected, she must have gotten a normal “A” from her mother (I-2). So her genotype is Aa → yes, we *can* determine it.

Wait — but the question says “Can you determine...” and the handwritten answer says “No”. Let me double-check.

Actually, looking again: II-4 is female, open circle → unaffected. Her dad is I-1 (affected, aa). So she MUST get “a” from him. To be unaffected, she must have “A” from mom → so genotype is definitely Aa. So we CAN determine it.

But maybe the worksheet expects “no” because sometimes in pedigrees you don’t know for sure? Wait — no, in this case, since dad is aa, and she’s unaffected, she has to be Aa. So answer should be YES.

Hmm… perhaps there's a mistake in the original handwriting? Or maybe I misread who II-4 is?

Let me label:

Generation I:
I-1 (male, shaded)
I-2 (female, open)

Generation II:
Left to right:
II-1 (male, shaded)
II-2 (female, open)
II-3 (male, open)
II-4 (female, open) ← this is the one
II-5 (male, open)
II-6 (female, open)

Yes, II-4 is daughter of I-1 and I-2. Dad is aa → gives her “a”. She’s unaffected → must have “A” from mom → genotype Aa. So we CAN determine it.

But the student wrote “No” — maybe they made a mistake? Or maybe the question is tricky?

Wait — actually, looking back at the problem statement: “Assume the trait is rare unless otherwise noted.” And also, in part D, they ask for probability that III-2 is carrier.

Maybe for part C, they mean “can you determine with certainty?” — and yes, we can.

But let’s move on and come back.

Part D: Probability that III-2 is a carrier.

Who is III-2? Looking at the chart: III-2 is child of II-5 and II-6. Both II-5 and II-6 are unaffected. They had an affected child (III-3), so both must be carriers (Aa).

So parents: Aa x Aa

Possible offspring:
- AA (unaffected, not carrier) → 25%
- Aa (unaffected, carrier) → 50%
- aa (affected) → 25%

III-2 is unaffected (open square), so we exclude the aa possibility.

Among unaffected children, probability of being carrier = P(Aa | not aa) = 50% / 75% = 2/3

So answer: 2/3

That matches the handwritten answer.

Back to part C: Maybe the question is phrased as “Can you determine the genotype of II-4?” and since II-4 is unaffected and has an affected parent, yes — she must be Aa. So answer should be YES.

But the student wrote “No” — perhaps error? Or maybe I’m missing something.

Wait — let’s check the actual image description again. In the user’s note, they circled “Dd” and “dd”, which suggests they’re using D/d instead of A/a. Doesn’t matter — same logic.

Perhaps for II-4, if we don’t know the mother’s status? But mother I-2 is unaffected, and since trait is rare, likely homozygous dominant? Not necessarily — she could be carrier.

Ah! Here’s the catch: For II-4, we know dad is aa, so she got “a” from him. She is unaffected, so she must have at least one “A”. But do we know if she’s AA or Aa? No — wait, no: if she were AA, she’d have to get “A” from both parents. But dad is aa, so he can only give “a”. Therefore, she MUST be Aa. There’s no other possibility.

So genotype is determined: Aa.

Therefore, answer to C should be YES.

But maybe the worksheet considers that we don’t know if the mother contributed A or a? But mother is unaffected — if trait is recessive, mother could be AA or Aa. But for II-4, regardless of what mother gave, since dad gave “a”, and II-4 is unaffected, she must have gotten “A” from mother → so genotype is Aa.

I think the correct answer is YES.

However, since the student wrote “No”, and perhaps the teacher expects that, I’ll go with the logic.

Wait — let’s see the second problem.

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Problem 2: Second Pedigree (bottom one)

Again, shaded = affected.

Look at generation I:
- I-1 (male, shaded)
- I-2 (female, open)

Their children:
- II-1 (male, open)
- II-2 (female, shaded)
- II-3 (male, open)
- II-4 (female, open)
- II-5 (male, shaded)
- II-6 (female, open)

Then generation III:
From II-1 and II-2:
- III-1 (male, open)
- III-2 (female, shaded)
- III-3 (male, open)
- III-4 (female, shaded)

Also, II-5 and II-6 have children:
- III-5 (male, shaded)
- III-6 (female, open)
- III-7 (male, shaded)

Key observation: Affected individuals appear in every generation. Also, affected males have affected daughters and sons. And unaffected parents do NOT have affected children — except wait, look at II-1 and II-2: II-1 is unaffected, II-2 is affected, and they have affected children (III-2 and III-4). That’s fine.

But more importantly: Is there any case where two unaffected parents have an affected child? Let’s see:

- II-3 (unaffected male) and II-4 (unaffected female) — their children? Not shown, or maybe not present.

Actually, in this pedigree, all affected individuals have at least one affected parent. For example:
- I-1 affected → passes to II-2 and II-5
- II-2 affected → passes to III-2 and III-4
- II-5 affected → passes to III-5 and III-7

And no unaffected parents have affected children. That suggests the trait is dominant.

Because for dominant traits, you only need one copy to be affected, and affected individuals usually have an affected parent (unless new mutation, but we assume not).

Also, if it were recessive, we’d expect unaffected parents to sometimes have affected kids, but here it doesn’t happen.

So answer to part A: The trait is dominant.

Part B: Genotype of I-1.

If dominant, let’s use D for dominant (disease), d for normal.

I-1 is affected → could be DD or Dd.

But his wife I-2 is unaffected → must be dd.

Their children: some affected, some not.

For example, II-1 is unaffected → must be dd. He got one allele from each parent. From mom (dd) he got d. From dad (I-1) he must have gotten d → so I-1 must have given d to II-1. But I-1 is affected, so if he gave d, he must also have D → so genotype is Dd.

Similarly, II-2 is affected → got D from dad, d from mom → Dd.

So I-1 is Dd.

Answer: Dd

Part C: Can we determine genotype of II-4?

II-4 is unaffected female → so must be dd (since if she had D, she’d be affected).

Her parents: I-1 (Dd) and I-2 (dd). So she could be Dd or dd. But since she’s unaffected, she must be dd. Yes, we can determine it.

Handwritten answer says “No” — again, discrepancy.

But logically: unaffected → for dominant trait, must be homozygous recessive → dd. So yes, determined.

Part D: Probability that III-2 is heterozygous.

III-2 is affected female. Parents: II-1 (unaffected → dd) and II-2 (affected → we said Dd, since her dad was Dd and mom dd).

So cross: dd (dad) x Dd (mom)

Offspring:
- Dd (affected) → 50%
- dd (unaffected) → 50%

III-2 is affected → so she must be Dd. There’s no other possibility, because dad is dd, so he gives d; mom gives D or d. If III-2 is affected, she got D from mom → so genotype Dd.

Thus, probability she is heterozygous = 100% → 1

But handwritten answer says “1/2” — that would be if we didn’t know she’s affected, but we do.

Wait, perhaps I misidentified III-2.

In the bottom pedigree:

Generation III, left side: children of II-1 and II-2:

III-1: male, open → unaffected

III-2: female, shaded → affected

III-3: male, open → unaffected

III-4: female, shaded → affected

Yes, III-2 is affected.

Parents: II-1 (dd), II-2 (Dd)

So affected child must be Dd.

So probability = 1.

But maybe the question is different? Or perhaps in some interpretations...

Another thought: if the trait is dominant, and III-2 is affected, and one parent is unaffected (dd), then yes, she must be heterozygous.

So answer should be 1.

But let’s confirm with the other branch.

II-5 is affected male. His parents: I-1 (Dd) and I-2 (dd), so II-5 is Dd.

He marries II-6 (unaffected → dd)

Their children: III-5 (affected), III-6 (unaffected), III-7 (affected)

Each child: 50% chance Dd (affected), 50% dd (unaffected)

III-5 is affected → so Dd

III-7 is affected → Dd

III-6 unaffected → dd

All consistent.

So for III-2, same thing: affected, with unaffected father → must be Dd.

So probability = 1.

But perhaps the worksheet has a typo, or I’m misreading.

Maybe for part D, it’s asking for something else.

Looking back at user’s note: for first problem, D is 2/3, which is correct.

For second problem, D is written as "1/2" — but according to logic, it should be 1.

Unless... is III-2 not necessarily from those parents? No, the lines connect.

Perhaps in some pedigrees, if there’s consanguinity or something, but not here.

Another idea: maybe for dominant traits, if an individual is affected and has an unaffected parent, they must be heterozygous, which is true.

So I think the correct answers are:

Problem 1:
A: recessive
B: dd (or aa)
C: Yes (genotype is Dd or Aa)
D: 2/3

Problem 2:
A: dominant
B: Dd
C: Yes (genotype is dd)
D: 1

But since the student has different answers, and perhaps the worksheet expects certain responses, I need to go with standard genetics rules.

Perhaps for problem 2 part C, "can you determine genotype of II-4" — II-4 is unaffected, so for dominant trait, she must be dd, so yes.

But let’s see the handwritten answers in the image description: for problem 2, C is "No", D is "1/2"

Why would D be 1/2? Only if we didn't know III-2's phenotype, but we do.

Unless the question is: "what is the probability that III-2 is heterozygous given that she is affected?" — which is 1.

Or perhaps it's a trick.

Another possibility: in some cases, if the trait is dominant but lethal in homozygotes, but not indicated.

I think based on standard interpretation, my reasoning is correct.

But to match common textbook problems, let's assume that for problem 2 part D, if III-2 is affected, and parents are dd and Dd, then she is definitely Dd, so probability 1.

Perhaps the "1/2" is for a different individual.

Let's calculate for III-6: unaffected, parents II-5 (Dd) and II-6 (dd), so III-6 is dd, probability of being heterozygous is 0.

Not helpful.

Or for III-1: unaffected, parents II-1 (dd) and II-2 (Dd), so III-1 is dd, not heterozygous.

I think there might be a mistake in the expected answer.

Perhaps for problem 2 part D, it's asking for the probability that a child of II-5 and II-6 is heterozygous, but it specifies III-2.

III-2 is from II-1 and II-2.

I think I have to go with logic.

So final answers:

Problem 1:
A: recessive
B: dd
C: Yes
D: 2/3

Problem 2:
A: dominant
B: Dd
C: Yes
D: 1

But since the user might expect the handwritten answers, and to avoid confusion, I'll provide the correct genetic answers.

Perhaps in problem 2, for part C, "can you determine genotype of II-4" — II-4 is unaffected, so for dominant trait, she is dd, so yes.

But let's box the answers as per standard.

After careful thought, I recall that in some pedigrees, if an individual is unaffected and the trait is dominant, their genotype is known to be homozygous recessive.

So for II-4 in problem 2, genotype is dd, determined.

Similarly for others.

So I'll proceed.

Final Answer:

For Problem 1:
A: recessive
B: dd
C: Yes
D: \frac{2}{3}

For Problem 2:
A: dominant
B: Dd
C: Yes
D: 1

But to write it neatly.