Pedigree chart practice worksheet for understanding human genetic disorders, focusing on albinism inheritance.

Let’s solve this step by step.

We’re told that albinism is a recessive trait. That means:

- To have albinism (the disorder), an animal must have two copies of the recessive allele → genotype = aa
- If an animal has at least one dominant allele (A), it will look normal → genotypes AA or Aa both give “normal” appearance

In pedigrees:
- White symbols = normal phenotype
- Gray/shaded symbols = albino (affected)

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Part 1: First Pedigree

Top generation: two white circles/squares → both are parents, and they had children including some shaded ones.

Their children:
- One shaded square (albino male)
- Two white circles (normal females)
→ So total 3 children

The oldest child? In pedigree charts, we usually read left to right as birth order. The first child shown is the white circle on the far left → so female

Now, how many grandchildren?

Look at the next generation:

- The albino son (shaded square) married a white circle → they had one child: a shaded circle (albino daughter) → that’s 1 grandchild from him
- The middle white circle (daughter) married a white square → they had two children: one white circle and one shaded circle → that’s 2 more grandchildren
- The rightmost white circle (daughter) — no spouse or children shown → so 0 grandchildren from her

Total grandchildren = 1 + 2 + 0 = 3

Wait — let me double-check the diagram structure.

Actually, looking again:

First generation: couple (white circle + white square)

They have 3 children (left to right):
1. White circle (female)
2. Shaded square (male, albino)
3. White circle (female)

Then:

Child #1 (white circle) marries someone not shown? No — actually, in standard pedigree notation, if there’s a line connecting to another symbol, that’s the spouse.

Looking carefully:

- Child #2 (shaded square) is connected to a white circle → their child is a shaded circle → 1 grandchild
- Child #3 (white circle) is connected to a white square → their children are: one white circle and one shaded circle → 2 grandchildren
- Child #1 (white circle) — no spouse or children drawn → so 0

So yes, total grandchildren = 1 + 2 = 3

But wait — the question says “how many grandchildren does the couple have?” meaning the original top couple.

Yes — all those kids’ kids are their grandchildren.

So:

a) How many children? → 3
b) Sex of oldest child? → female (first child shown, leftmost)
c) How many grandchildren? → 3

Now fill in blanks for genotypes.

Original parents (top row): both white (normal), but they had an albino child (aa). That means BOTH parents must carry the recessive allele → so both are Aa

Because only way to get aa child is if each parent gives an ‘a’.

So:

Top left circle (mother): Aa
Top right square (father): Aa

Their children:

- Left child (white circle): could be AA or Aa — but since she has no affected children yet, we can’t tell for sure — BUT in part 2, we’ll see she might be carrier. For now, leave blank? Wait — the instruction says “fill out the blanks on the pedigree” — probably referring to labeling phenotypes under genotypes given.

Wait — re-read:

“Given the following genotypes, describe the phenotypes: (normal or albino)”

AA = ?
Aa = ?
aa = ?

That’s straightforward:

- AA → normal (has dominant allele)
- Aa → normal (still has dominant allele, just carrier)
- aa → albino (no dominant allele)

So:

AA = normal
Aa = normal
aa = albino

Then “Fill out the blanks on the pedigree” — likely means label each person with their genotype based on what we know.

Let’s do that.

Generation I (top):

Both parents are normal but had albino child → so both must be carriers → Aa and Aa

Generation II (their children):

Left child: white circle → normal → could be AA or Aa. But later, in Generation III, she has no children shown, so we don’t know. Actually, wait — in the diagram, is she connected to anyone? Looking back — in my earlier analysis, I think I misread.

Actually, let me reconstruct the pedigree clearly.

Standard layout:

Row 1: Parents — Circle (female) and Square (male), both white → Genotype: both Aa (since they produced aa child)

Row 2: Their offspring — three individuals:

- Left: Circle (female), white → let’s call her II-1
- Middle: Square (male), shaded → II-2 → must be aa
- Right: Circle (female), white → II-3

Now, II-2 (aa male) is married to a white circle (II-2’s wife) → let’s call her II-4 (not blood relative, so we don’t know her genotype yet). They have one child: shaded circle → III-1 → aa

Since III-1 is aa, she got one ‘a’ from dad (who is aa, so he gave ‘a’) and one ‘a’ from mom → so mom (II-4) must also be carrier → Aa

Similarly, II-3 (white circle) is married to a white square (II-5) → they have two children:

- III-2: white circle → normal
- III-3: shaded circle → aa

So III-3 is aa → got one ‘a’ from each parent → so both II-3 and II-5 must be carriers → Aa

Now, what about II-1? She’s white, no children shown → so we don’t know if she’s AA or Aa. But since the question doesn’t ask us to assign genotypes to everyone, maybe we only need to answer the questions asked.

Back to questions:

a) How many children does the couple have? → The top couple (I-1 and I-2) have 3 children → 3

b) What is the sex of the oldest child? → Assuming left to right is birth order, first child is female → female

c) How many grandchildren does the couple have? → Children of their children:

- From II-2 and his wife: 1 child (III-1)
- From II-3 and her husband: 2 children (III-2 and III-3)
- From II-1: 0 children shown

Total = 1 + 2 = 3

Now, for the genotype descriptions:

AA = normal
Aa = normal
aa = albino

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Part 2: Second Pedigree

This one starts with a white circle and white square (generation I) → they have children:

- One shaded square (albino male)
- One white square (normal male)
- One shaded circle (albino female)

So three children.

Then, the shaded square (II-1) marries a white circle (II-2) → they have two children: one shaded circle and one white circle

The shaded circle (II-3) marries a white square (II-4) → they have two children: one shaded circle and one white circle

Questions:

a) How many children did the original couple have? → 3 (as above)

b) How many grandchildren? →

From II-1 and II-2: 2 children
From II-3 and II-4: 2 children
Total = 4

c) How many of their grandchildren are albino? →

From II-1 & II-2: one shaded (albino), one white → 1 albino
From II-3 & II-4: one shaded (albino), one white → 1 albino
Total albino grandchildren = 2

d) Create an extension... showing other grandparents marrying and having a son who is NOT albino.

We need to add parents for the spouses (II-2 and II-4), since they are not related to the original family.

For example:

Add a new couple (say, III-4 and III-5) who are parents of II-2 (the wife of II-1). Make them both normal, and have them produce a son (II-2’s brother?) who is not albino — which is easy, since most people aren’t albino.

But specifically: “show the other grandparents marrying and having a son who is NOT albino”

So perhaps: draw parents for II-2 (who is currently just a white circle with no parents shown). Let’s say we add:

New generation above II-2: a couple (both white) → they have a son (who is II-2’s brother? Or maybe II-2 herself is female, so her sibling is male).

To keep it simple: Add two new symbols (circle and square, both white) connected as parents, and they have a son (square, white) — that son is not albino, as required.

And connect that son appropriately — but since II-2 is already married, maybe make the new son a sibling of II-2.

But the question says “create an extension to the pedigree to show the other grandparents marrying and having a son who is NOT albino.”

So “other grandparents” likely refers to the parents of the spouses (II-2 and II-4).

So for example, take II-2 (wife of II-1). Her parents are not shown. Add them: a couple (both normal), and they have a son (besides II-2) who is normal (not albino).

Similarly for II-4.

But to satisfy the question, adding one such pair is enough.

So in drawing terms: above II-2, add a circle and square (parents), connect them, then connect to II-2 and also to a new square (son) who is white (normal).

Same for II-4 if needed, but one instance suffices.

Now, back to answering.

Final answers:

Part 1:

Genotype to phenotype:

AA = normal
Aa = normal
aa = albino

Pedigree questions:

a) 3
b) female
c) 3

Part 2:

a) 3
b) 4
c) 2
d) [description of added pedigree part]

For d, since we can't draw, we describe:

Add a new set of parents (both normal) above one of the spouses (e.g., the wife of the albino male in generation II). Connect them to that spouse and also to a new son (normal, white square) to show that these grandparents had a non-albino son.

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Now, compiling final answer.

Note: The user didn’t specify which part to focus on, but since both parts are included, I’ll answer all.

Also, for the genotype blanks in part 1, we filled:

AA = normal
Aa = normal
aa = albino

And for the pedigree filling — though not explicitly asked beyond the questions, we used logic to find numbers.

I think we’re ready.

Final Answer:

Part 1:
- AA = normal
- Aa = normal
- aa = albino
a) 3
b) female
c) 3

Part 2:
a) 3
b) 4
c) 2
d) Add a new couple (both normal) as parents of one of the spouses (e.g., the wife of the albino male in generation II). Show they have a son (normal, white square) in addition to that spouse.