complete the table below by matching the correct blood type to the offspring.
| | $i^a$ | $i^b$ | $i$ |
|-------|---------------------|---------------------|---------------------|
| $i^a$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |
| $i^b$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |
| $i$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |

Question

complete the table below by matching the correct blood type to the offspring.
|       | $i^a$               | $i^b$               | $i$                 |
|-------|---------------------|---------------------|---------------------|
| $i^a$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |
| $i^b$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |
| $i$   | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ | select choice $\boldsymbol{\downarrow}$ |

Accepted Answer

# Explanation:
To solve this, we use the principles of blood type genetics (ABO system, where $ I^A $ and $ I^B $ are codominant, $ i $ is recessive). Let's analyze each cell:

## Step 1: $ I^A \times I^A $
Possible genotypes: $ I^A I^A $ or $ I^A i $ (since both parents give $ I^A $ or one gives $ I^A $, one $ i $ – but wait, if both are $ I^A $, the offspring can be $ I^A I^A $ (type A) or $ I^A i $ (type A). Wait, actually, if both parents are $ I^A $ (homozygous or heterozygous? Wait, the table is about alleles. Wait, the rows and columns are alleles: $ I^A $, $ I^B $, $ i $. So each cell is the combination of one allele from the row (parent 1) and one from the column (parent 2).

So for $ I^A $ (row) and $ I^A $ (column): offspring genotypes are $ I^A I^A $ (type A) or $ I^A I^A $ (wait, no: each parent contributes one allele. So parent 1: $ I^A $, parent 2: $ I^A $ → offspring: $ I^A I^A $ (type A) or if one is $ I^A $ and the other $ i $, but the table is about the alleles, not the parents' genotypes. Wait, the table is structured as: rows = allele from parent 1, columns = allele from parent 2. So each cell is the combination of the two alleles.

So:

- $ I^A \times I^A $: alleles $ I^A $ and $ I^A $ → genotype $ I^A I^A $ → blood type A.
- $ I^A \times I^B $: alleles $ I^A $ and $ I^B $ → genotype $ I^A I^B $ → blood type AB.
- $ I^A \times i $: alleles $ I^A $ and $ i $ → genotype $ I^A i $ → blood type A.
- $ I^B \times I^A $: same as $ I^A \times I^B $ → $ I^A I^B $ → AB.
- $ I^B \times I^B $: $ I^B I^B $ → blood type B.
- $ I^B \times i $: $ I^B i $ → blood type B.
- $ i \times I^A $: $ I^A i $ → blood type A.
- $ i \times I^B $: $ I^B i $ → blood type B.
- $ i \times i $: $ ii $ → blood type O.

Let's fill each cell:

1. $ I^A $ (row) × $ I^A $ (column): A (genotype $ I^A I^A $ or $ I^A i $? Wait, no: the alleles are $ I^A $ and $ I^A $, so the offspring's genotype is $ I^A I^A $ (type A) or if one parent is $ I^A $ and the other $ i $, but the table is about the alleles, so each cell is the combination of the two alleles. Wait, actually, the table is showing the possible offspring blood type when one parent contributes the row allele and the other the column allele. So:

- $ I^A $ (row) & $ I^A $ (column): offspring gets $ I^A $ from both → genotype $ I^A I^A $ → blood type A.
- $ I^A $ & $ I^B $: $ I^A I^B $ → AB.
- $ I^A $ & $ i $: $ I^A i $ → A.
- $ I^B $ & $ I^A $: $ I^A I^B $ → AB.
- $ I^B $ & $ I^B $: $ I^B I^B $ → B.
- $ I^B $ & $ i $: $ I^B i $ → B.
- $ i $ & $ I^A $: $ I^A i $ → A.
- $ i $ & $ I^B $: $ I^B i $ → B.
- $ i $ & $ i $: $ ii $ → O.

Now let's list each cell:

- Row $ I^A $, Column $ I^A $: Type A
- Row $ I^A $, Column $ I^B $: Type AB
- Row $ I^A $, Column $ i $: Type A
- Row $ I^B $, Column $ I^A $: Type AB
- Row $ I^B $, Column $ I^B $: Type B
- Row $ I^B $, Column $ i $: Type B
- Row $ i $, Column $ I^A $: Type A
- Row $ i $, Column $ I^B $: Type B
- Row $ i $, Column $ i $: Type O

Wait, but the table has dropdowns. Let's confirm each:

1. $ I^A $ (row) × $ I^A $ (column): Offspring alleles: $ I^A $ (from row) and $ I^A $ (from column) → genotype $ I^A I^A $ → blood type A.
2. $ I^A $ × $ I^B $: $ I^A I^B $ → AB.
3. $ I^A $ × $ i $: $ I^A i $ → A.
4. $ I^B $ × $ I^A $: $ I^A I^B $ → AB.
5. $ I^B $ × $ I^B $: $ I^B I^B $ → B.
6. $ I^B $ × $ i $: $ I^B i $ → B.
7. $ i $ × $ I^A $: $ I^A i $ → A.
8. $ i $ × $ I^B $: $ I^B i $ → B.
9. $ i $ × $ i $: $ ii $ → O.

So each cell's blood type:

- $ I^A $-$ I^A $: A
- $ I^A $-$ I^B $: AB
- $ I^A $-$ i $: A
- $ I^B $-$ I^A $: AB
- $ I^B $-$ I^B $: B
- $ I^B $-$ i $: B
- $ i $-$ I^A $: A
- $ i $-$ I^B $: B
- $ i $-$ i $: O

# Answer:
To complete the table, the blood types for each cell (row × column) are:

- $ \boldsymbol{I^A \times I^A} $: Type A
- $ \boldsymbol{I^A \times I^B} $: Type AB
- $ \boldsymbol{I^A \times i} $: Type A
- $ \boldsymbol{I^B \times I^A} $: Type AB
- $ \boldsymbol{I^B \times I^B} $: Type B
- $ \boldsymbol{I^B \times i} $: Type B
- $ \boldsymbol{i \times I^A} $: Type A
- $ \boldsymbol{i \times I^B} $: Type B
- $ \boldsymbol{i \times i} $: Type O

(If the dropdowns have options like "A", "B", "AB", "O", select the corresponding one for each cell using the above logic.)

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QUESTION IMAGE

Question

Explanation:

Step 1: \( I^A \times I^A \)

Answer:

	$i^a$	$i^b$	$i$
$i^b$	select choice $\boldsymbol{\downarrow}$	select choice $\boldsymbol{\downarrow}$	select choice $\boldsymbol{\downarrow}$
$i$	select choice $\boldsymbol{\downarrow}$	select choice $\boldsymbol{\downarrow}$	select choice $\boldsymbol{\downarrow}$