Solved Pedigree Worksheet pedigree is a chart of a persons ... - Free Printable

1.
a. I-1: Dd
b. I-2: dd
c. I-3: Dd
d. I-4: dd
e. II-1: Dd
f. II-2: dd
g. II-3: Dd
h. II-4: dd
i. II-5: Dd
j. III-1: Dd
k. III-2: dd
l. III-3: Dd

2.
a. III-3: heterozygous
b. I-1: heterozygous
c. II-1: heterozygous
d. II-4: homozygous recessive

3. This trait is autosomal dominant. We know this because:
- The trait appears in every generation (it is not skipping generations).
- Affected individuals have at least one affected parent.
- Both males and females are affected, and the trait is not sex-linked.
- The affected individuals (e.g., I-1, II-1, II-3, III-1) are often heterozygous, and unaffected individuals are homozygous recessive.

4. We can know for sure that individuals II-3 and II-4 are heterozygous because:
- II-3 is affected (shaded) and has an unaffected child (III-2), so II-3 must carry the recessive allele (d) to pass it on. Therefore, II-3 must be Dd.
- II-4 is unaffected (unshaded) and has an affected child (III-1), so II-4 must carry the recessive allele (d). But since II-4 is unaffected, II-4 must be dd. Wait — this contradicts the question. Let's recheck.
Actually, II-4 is unaffected (unshaded), so II-4 must be dd. But the question says "how can you know for sure that individuals II-3 and II-4 are heterozygous?" — but II-4 is unaffected, so cannot be heterozygous dominant.
Wait — rechecking the pedigree: II-4 is unshaded, so unaffected. Therefore, II-4 is dd (homozygous recessive).
But II-3 is shaded (affected), and has a child III-1 (shaded), but also has a child III-2 (unshaded). So II-3 must be Dd (heterozygous), because if II-3 were DD, all children would be affected. Since III-2 is unaffected, II-3 must be Dd.
II-4 is unaffected, so II-4 is dd. So II-4 is not heterozygous.
But the question says "II-3 and II-4 are heterozygous" — but II-4 is not. So perhaps there's a mistake in the question or the pedigree labeling.
Wait — perhaps II-4 is affected? No, II-4 is unshaded.
Wait — let's re-express:
II-3 is affected (shaded), and has an unaffected child (III-2). So II-3 must be heterozygous (Dd).
II-4 is unaffected (unshaded), so II-4 must be dd. So II-4 is homozygous recessive, not heterozygous.
Therefore, the question must have a typo, or I misread.
But the question says "how can you know for sure that individuals II-3 and II-4 are heterozygous?" — but II-4 cannot be heterozygous because they are unaffected.
Unless the trait is recessive? But earlier we said it's dominant.
Wait — perhaps the trait is recessive? But the first question says "autosomal dominant".
So if the trait is autosomal dominant, then unaffected individuals must be dd.
So II-4 is unaffected → dd → homozygous recessive → not heterozygous.
Therefore, the question is flawed, or there's a misunderstanding.
But perhaps the pedigree is different.
Wait — in the first pedigree, II-4 is unshaded, so unaffected.
So II-4 cannot be heterozygous for a dominant trait.
So the only way II-4 could be heterozygous is if the trait is recessive.
But the first question says "autosomal dominant".
So contradiction.
Therefore, the question must be wrong, or I misread.
But perhaps the second pedigree is different?
Wait — the user uploaded two pedigrees.
The first one is for questions 1-1, and the second one is for questions 2-4.
So the second pedigree is separate.
Let me recheck:
In the second pedigree:
I-1: shaded (affected male)
I-2: unshaded (unaffected female)
II-1: unshaded
II-2: unshaded
II-3: shaded (affected female)
II-4: shaded (affected male)
II-5: unshaded
II-6: unshaded
II-7: unshaded
II-8: unshaded
III-1: unshaded
III-2: unshaded
III-3: shaded
III-4: shaded
III-5: unshaded
III-6: unshaded
III-7: unshaded
III-8: unshaded
So II-3 is shaded (affected), II-4 is shaded (affected).
They have children: III-3 and III-4 are shaded (affected), III-5 is unshaded (unaffected).
So II-3 and II-4 are both affected, but they have an unaffected child (III-5).
Therefore, both II-3 and II-4 must be heterozygous (Dd), because if either were DD, all children would be affected. Since they have an unaffected child (dd), both must be Dd.
So yes, II-3 and II-4 are heterozygous.
So the answer is:
We can know for sure that II-3 and II-4 are heterozygous because they are both affected (shaded) but have an unaffected child (III-5), which must be dd. Therefore, both parents must carry the recessive allele (d), so they must be Dd.