blood-type-codominance-practice-problems.pdf - Blood Type ... - Free Printable

Let’s solve each problem step by step.

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Problem 1: Fill in the table for blood type genotypes and phenotypes.

We’re given some info and need to fill in the blanks.

Recall:
- Blood types are determined by three alleles: I^A, I^B, and i.
- I^A and I^B are codominant; i is recessive.
- Phenotype = what you see (blood type A, B, AB, or O)
- Genotype = the actual gene combination

Given table:

| Genotype | Phenotype | Can donate to | Can receive from |
|----------------|---------------|----------------------|------------------------|
| ii | O | Everyone (universal donor) | Only O |
| I^A I^A or I^A i | A | A, AB | A, O |
| I^B I^B or I^B i | B | B, AB | B, O |
| I^A I^B | AB | AB only | Everyone (universal recipient) |

Now let’s match with what’s written in the image (handwritten answers):

Row 1: Genotype “ii” → Phenotype “O” → Donate to “Everyone” → Receive from “O” ✔ Correct

Row 2: Genotype “I^A I^A or I^A i” → Phenotype “A” → Donate to “A, AB” → Receive from “A, O” ✔ Correct

Row 3: Genotype “I^B I^B or I^B i” → Phenotype “B” → Donate to “B, AB” → Receive from “B, O” ✔ Correct

Row 4: Genotype “I^A I^B” → Phenotype “AB” → Donate to “AB” → Receive from “Everyone” ✔ Correct

So the table is already correctly filled out in the image. No changes needed.

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Problem 2: Write the genotype for each person based on the description.

a) Homozygous for the “B” allele → That means two B alleles → I^B I^B

b) Heterozygous for the “A” allele → One A and one other (must be i, since if it were B, it’d be AB) → I^A i

c) Type O → Must be ii → ii

d) Type “A” and had a type “O” parent → To have an O parent (ii), you must have gotten an i from them. So your genotype must be I^A i → I^A i

e) Type “AB” → Only possible genotype → I^A I^B

f) Blood can be donated to anybody → Universal donor → Type O → ii

g) Can only get blood from a type “O” donor → That means they can’t have any A or B antigens → Must be type O → ii

All handwritten answers match these → All correct.

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Problem 3: If a man who is heterozygous for the “B” allele (I^B i) marries a woman who is type “AB” (I^A I^B), what percentage of their children will have each blood type?

Set up Punnett square:

Man: I^B i
Woman: I^A I^B

Gametes from man: I^B or i
Gametes from woman: I^A or I^B

Punnett Square:

```
| I^A | I^B
-----------------------------
I^B | I^A I^B | I^B I^B
i | I^A i | I^B i
```

Offspring genotypes:
- I^A I^B → AB
- I^B I^B → BB → phenotype B
- I^A i → Ai → phenotype A
- I^B i → Bi → phenotype B

So phenotypes:
- AB: 1 out of 4 → 25%
- B: 2 out of 4 → 50% (I^B I^B and I^B i)
- A: 1 out of 4 → 25%

Handwritten answer says:
Type A: 25%
Type B: 50%
Type AB: 25%
Type O: 0% → Correct! (No ii possible)

✔ All correct.

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Problem 4: Two parents think their baby was switched at the hospital. Mother is type “A”, father is type “B”, baby is type “O”. Was the baby switched?

First, figure out possible genotypes.

Mother: Type A → could be I^A I^A or I^A i
Father: Type B → could be I^B I^B or I^B i
Baby: Type O → must be ii

For baby to be ii, both parents must contribute an i allele.

So mother must be I^A i (not I^A I^A)
Father must be I^B i (not I^B I^B)

Is that possible? Yes!

Punnett square:

Mother: I^A i
Father: I^B i

Gametes:
Mom: I^A or i
Dad: I^B or i

Offspring:
- I^A I^B → AB
- I^A i → A
- I^B i → B
- i i → O ← This is the baby!

So yes — it’s genetically possible for two parents (one type A, one type B) to have a type O child — IF both are heterozygous.

Therefore, the baby was NOT necessarily switched. It’s possible.

Handwritten answer says: “No, because both parents could carry the ‘i’ allele.” → Correct!

Also shows Punnett square with 25% chance of O → Correct.

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Final Check:

All problems solved correctly as per standard genetics rules.

Final Answer:
All answers provided in the image are correct.