Question

use the punnett square below to answer the next question.
taking into consideration all of the offspring, what is the probability that this couple will have a colorblind male?
0% (0 out of 4)
25% (1 out of 4)
50% (2 out of 4)
100% (4 out of 4)

Explanation:

Step1: Analyze Punnett square

The Punnett square has 4 cells (offspring possibilities). For a colorblind male, the genotype is \(X^bY\). Let's fill the Punnett square:
- Top - left: \(X \times X^b = X X^b\) (female, not colorblind)
- Top - right: \(X^b \times X^b = X^b X^b\) (female, colorblind)
- Bottom - left: \(X \times Y = X Y\) (male, not colorblind)
- Bottom - right: \(X^b \times Y = X^b Y\) (male, colorblind)

Wait, no, wait. Wait the rows are the father's gametes? Wait, actually, the columns are one parent (say mother) with \(X\) and \(X^b\), and rows are father with \(X^b\) and \(Y\)? Wait no, maybe the mother is \(X X^b\) (carrier) and father is \(X^b Y\)? Wait no, let's re - evaluate. Wait the Punnett square:

First, let's correctly determine the genotypes. The possible offspring genotypes:

1. \(X^b\) (row) and \(X\) (column): \(X X^b\) (female, carrier)
2. \(X^b\) (row) and \(X^b\) (column): \(X^b X^b\) (female, colorblind)
3. \(Y\) (row) and \(X\) (column): \(X Y\) (male, normal)
4. \(Y\) (row) and \(X^b\) (column): \(X^b Y\) (male, colorblind)

Wait, no, the number of male offspring: out of 4 offspring, 2 are male (the ones with Y). Among the male offspring, 1 is \(X^b Y\) (colorblind) and 1 is \(X Y\) (normal). Wait, no, let's count the total number of offspring. The Punnett square has 4 cells. The male offspring are the two in the row with Y. Among them, one is \(X^b Y\) (colorblind) and one is \(X Y\) (normal). Wait, but the question is about all offspring (4 total). The colorblind male is \(X^b Y\), which is 1 out of 4? No, wait no. Wait let's re - do the Punnett square filling:

- Cell 1: \(X^b\) (row) and \(X\) (column): Genotype \(X X^b\) (female)
- Cell 2: \(X^b\) (row) and \(X^b\) (column): Genotype \(X^b X^b\) (female)
- Cell 3: \(Y\) (row) and \(X\) (column): Genotype \(X Y\) (male)
- Cell 4: \(Y\) (row) and \(X^b\) (column): Genotype \(X^b Y\) (male)

Now, the colorblind male is \(X^b Y\), which is 1 out of 4? No, wait the male offspring are 2 (cells 3 and 4). The colorblind male is 1 (cell 4). But the question says "taking into consideration ALL of the offspring". So total offspring = 4. The number of colorblind males is 1? Wait no, wait maybe I made a mistake. Wait, let's check again.

Wait, the father's gametes: if the father is \(X^b Y\), his gametes are \(X^b\) and \(Y\). The mother's gametes: if the mother is \(X X^b\), her gametes are \(X\) and \(X^b\). Then the Punnett square:

| | \(X\) | \(X^b\) |
|---|---|---|
| \(X^b\) | \(X X^b\) | \(X^b X^b\) |
| \(Y\) | \(X Y\) | \(X^b Y\) |

Now, the offspring:

- \(X X^b\): female, carrier
- \(X^b X^b\): female, colorblind
- \(X Y\): male, normal
- \(X^b Y\): male, colorblind

So total offspring: 4. Number of colorblind males: 1? Wait no, \(X^b Y\) is 1. Wait but the options have 25% (1 out of 4), 50% (2 out of 4), etc. Wait, maybe the mother is \(X^b X^b\) and father is \(X Y\)? No, let's re - examine the Punnett square given. The rows are \(X^b\) and \(Y\), columns are \(X\) and \(X^b\).

Wait, let's list all 4 offspring:

1. \(X^b\) (row) and \(X\) (column): \(X X^b\) (female)
2. \(X^b\) (row) and \(X^b\) (column): \(X^b X^b\) (female)
3. \(Y\) (row) and \(X\) (column): \(X Y\) (male)
4. \(Y\) (row) and \(X^b\) (column): \(X^b Y\) (male)

Now, the colorblind male is \(X^b Y\), which is 1 out of 4? No, wait the male offspring are 2 (rows with Y). Among the male offspring, 1 is colorblind (\(X^b Y\)) and 1 is normal (\(X Y\)). But the question is about all offspring (4). So the number of colorblind males is 1. Wait, but 1 out of 4 is 25%? No, wait no. Wait, maybe I messed up the parents' genotypes. Wait, if the mother is \(X X^b\) (carrier) and father is \(X^b Y\) (colorblind), then:

Mother's gametes: \(X\), \(X^b\)

Father's gametes: \(X^b\), \(Y\)

Punnett square:

| | \(X\) | \(X^b\) |
|---|---|---|
| \(X^b\) | \(X X^b\) | \(X^b X^b\) |
| \(Y\) | \(X Y\) | \(X^b Y\) |

Offspring:

- \(X X^b\): female, carrier
- \(X^b X^b\): female, colorblind
- \(X Y\): male, normal
- \(X^b Y\): male, colorblind

So total offspring: 4. Number of colorblind males: 1? Wait, no, \(X^b Y\) is 1. Wait, but 1 out of 4 is 25%? But wait, the male offspring are 2 (the two with Y). So the probability of a colorblind male is (number of colorblind males) / (total number of offspring) = 1/4 = 25%? Wait no, wait \(X^b Y\) is 1, total offspring 4, so 1/4 = 25%? But wait, maybe the mother is \(X^b X^b\) and father is \(X Y\). Then:

Mother's gametes: \(X^b\), \(X^b\)

Father's gametes: \(X\), \(Y\)

Punnett square:

| | \(X^b\) | \(X^b\) |
|---|---|---|
| \(X\) | \(X X^b\) | \(X X^b\) |
| \(Y\) | \(X^b Y\) | \(X^b Y\) |

Then the offspring:

- \(X X^b\) (female, carrier)
- \(X X^b\) (female, carrier)
- \(X^b Y\) (male, colorblind)
- \(X^b Y\) (male, colorblind)

In this case, the number of colorblind males is 2, total offspring 4, so 2/4 = 50%. Wait, this is a problem. So which is it?

Wait the original Punnett square has rows \(X^b\) and \(Y\), columns \(X\) and \(X^b\). So the father's gametes are \(X^b\) and \(Y\), mother's are \(X\) and \(X^b\). So the first case (mother \(X X^b\), father \(X^b Y\)) gives 1 colorblind male, the second case (mother \(X^b X^b\), father \(X Y\)) gives 2 colorblind males.

Wait, the key is to look at the Punnett square as given. Let's fill in the genotypes:

1. Top - left: \(X^b\) (row) and \(X\) (column): \(X X^b\)
2. Top - right: \(X^b\) (row) and \(X^b\) (column): \(X^b X^b\)
3. Bottom - left: \(Y\) (row) and \(X\) (column): \(X Y\)
4. Bottom - right: \(Y\) (row) and \(X^b\) (column): \(X^b Y\)

Now, the colorblind male is \(X^b Y\), which is 1 out of 4? No, wait \(X^b Y\) is 1, total offspring 4, so 1/4 = 25%? But wait, if we consider that males are 2 (the two with Y), then the probability of a colorblind male is (number of colorblind males) / (total number of offspring) = 1/4 = 25%? Wait, no, the total number of offspring is 4. The number of colorblind males is 1. So 1/4 = 25% (1 out of 4). But wait, maybe I made a mistake. Wait, let's count again:

Offspring:

1. Female: \(X X^b\) (not colorblind male)
2. Female: \(X^b X^b\) (not colorblind male)
3. Male: \(X Y\) (not colorblind)
4. Male: \(X^b Y\) (colorblind)

So out of 4 offspring, 1 is a colorblind male. So the probability is 1/4 = 25% (1 out of 4). Wait, but the option 25% (1 out of 4) is there. Wait, but wait, maybe the parents are different. Wait, if the mother is \(X^b X^b\) (colorblind) and father is \(X Y\) (normal), then:

Mother's gametes: \(X^b\), \(X^b\)

Father's gametes: \(X\), \(Y\)

Punnett square:

| | \(X^b\) | \(X^b\) |
|---|---|---|
| \(X\) | \(X X^b\) | \(X X^b\) |
| \(Y\) | \(X^b Y\) | \(X^b Y\) |

Offspring:

- \(X X^b\) (female, carrier)
- \(X X^b\) (female, carrier)
- \(X^b Y\) (male, colorblind)
- \(X^b Y\) (male, colorblind)

In this case, number of colorblind males is 2, total offspring 4, so 2/4 = 50%. But the Punnett square given has columns \(X\) and \(X^b\), rows \(X^b\) and \(Y\), which matches the first case (mother \(X X^b\), father \(X^b Y\))? Wait no, father with \(X^b\) and \(Y\) would mean father is \(X^b Y\), so his gametes are \(X^b\) and \(Y\). Mother with \(X\) and \(X^b\) is \(X X^b\). So the first Punnett square.

Wait, the confusion is about the parents' genotypes. But based on the given Punnett square, let's list all 4 offspring:

1. \(X^b\) (row) & \(X\) (column): \(X X^b\) (female)
2. \(X^b\) (row) & \(X^b\) (column): \(X^b X^b\) (female)
3. \(Y\) (row) & \(X\) (column): \(X Y\) (male)
4. \(Y\) (row) & \(X^b\) (column): \(X^b Y\) (male)

So the colorblind male is \(X^b Y\), which is 1 out of 4 offspring. So the probability is 1/4 = 25% (1 out of 4).

Step2: Calculate probability

Probability = (Number of colorblind male offspring) / (Total number of offspring) = \( \frac{1}{4} = 25\% \) (1 out of 4)

Answer:

25% (1 out of 4)