Pedigree charts worksheet - Free Printable

Let’s go step by step through both pedigree problems. We’re told that albinism is a recessive trait, which means:

- You need two copies of the “a” allele (aa) to have albinism.
- If you have AA or Aa, you look normal — but if you’re Aa, you’re a carrier.

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Problem 1

We are given genotypes and asked for phenotypes:

- AA = normal → two dominant alleles → no albinism
- Aa = normal → one dominant, one recessive → still looks normal (carrier)
- aa = albino → two recessive → has albinism

Now, let’s look at the first pedigree.

The top row: circle (female) and square (male), both white → they are parents.

They have three children below them:
- Left: white circle → daughter, normal
- Middle: gray square → son, albino → so genotype must be aa
- Right: white circle → daughter, normal

Then, the rightmost daughter (white circle) marries a white square (normal male). They have one child: gray circle → granddaughter, albino → so she is aa

Now answer the questions:

#### a) How many children does the couple have?

Look at the first generation (top couple). The lines going down from them connect to three shapes: circle, square, circle → that’s 3 children.

✔ Answer: 3

#### b) What is the sex of the oldest child?

In pedigrees, we usually read left to right as oldest to youngest.

Leftmost child is a circle → female.

✔ Answer: Female

#### c) How many grandchildren does the couple have?

Grandchildren come from their children having kids.

Only the rightmost child (the daughter who married) has a child → one gray circle.

So only 1 grandchild.

✔ Answer: 1

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Problem 2

This pedigree starts with a white circle and white square (both normal) → original couple.

They have three children:
- Left: gray square → albino son → aa
- Middle: gray circle → albino daughter → aa
- Right: white square → normal son → could be AA or Aa

Then:

- The albino son (gray square) marries a white circle (normal woman). They have one child: white square → normal son → so he must be Aa (since dad is aa, mom must give him an A).

- The albino daughter (gray circle) doesn’t marry or have kids shown.

- The normal son (white square) marries a gray circle (albino woman). They have two children:
- Gray circle → albino daughter → aa
- White circle → normal daughter → must be Aa (mom is aa, so gave her ‘a’; dad must have given ‘A’)

Now answer:

#### a) How many children did the original couple have?

Three: gray square, gray circle, white square → ✔ 3

#### b) How many grandchildren?

From the albino son: 1 child (white square)

From the normal son: 2 children (gray circle + white circle)

Total = 1 + 2 = ✔ 3

#### c) How many of their grandchildren are albino?

Look at the grandchildren:

- From albino son: white square → normal → not albino
- From normal son: gray circle → albino, white circle → normal

So only 1 grandchild is albino.

✔ Answer: 1

#### d) Create an extension to show the albino granddaughter marrying and having a son who is NOT albino.

The albino granddaughter is the gray circle under the normal son and albino wife.

She is aa.

To have a son who is NOT albino, he must have at least one A allele.

That means the father must give him an A.

So, she marries someone who is either AA or Aa.

If she marries an AA man → all kids will be Aa → normal.

If she marries an Aa man → 50% chance kid is Aa (normal), 50% aa (albino).

We just need to show it’s possible → so draw:

- Connect the albino granddaughter (gray circle) to a new square (man). Make him white (so he’s normal — could be AA or Aa).
- Draw a line down to a new square (son) → make him white (normal).

You don’t need to label genotypes unless asked, but since part d says “create an extension”, just adding those shapes is enough.

But since this is text, I’ll describe what to draw:

→ Add a horizontal line from the albino granddaughter (gray circle on bottom right) to a new white square (her husband).

→ From that couple, draw a vertical line down to a new white square (their son).

That shows she had a non-albino son.

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## Final Answer:

Problem 1:

Genotype → Phenotype:
- AA = normal
- Aa = normal
- aa = albino

a) 3
b) Female
c) 1

Problem 2:

a) 3
b) 3
c) 1
d) Draw: Albino granddaughter (gray circle) connected to a normal male (white square), with a line down to a normal son (white square).