Question

what is the probability of the offspring having red - green color blindness for the cross that is represented by the punnett square?
0 percent
25 percent
50 percent
75 percent

Explanation:

Step1: Analyze Punnett Square

The Punnett square has four cells. Genotypes: \(X^BX^B\), \(X^BY\), \(X^bX^B\), \(X^bY\). Red - green color blindness is X - linked recessive, so only \(X^bY\) (males) show it.

Step2: Calculate Probability

Number of \(X^bY\) (affected) is 1 out of 4. \(\frac{1}{4}= 25\%\)? Wait, no: Wait, \(X^bY\) is 1, total 4. But wait, color blindness in X - linked: females need \(X^bX^b\) to be blind, males \(X^bY\). In the square, the genotypes are: top - left \(X^BX^B\) (female, normal), top - right \(X^BY\) (male, normal), bottom - left \(X^bX^B\) (female, carrier), bottom - right \(X^bY\) (male, color blind). So 1 out of 4 offspring is color blind. Wait, 1/4 is 25%? Wait, no, 1 out of 4 is 25%? Wait, 4 cells: 1 affected. So probability is 25%? Wait, no, wait: 1 out of 4 is 25%? Wait, 4 offspring: 1 with \(X^bY\) (color blind). So 1/4 = 25%? Wait, but let's re - check. The Punnett square: rows are \(X^B\) and \(X^b\) (mother's gametes), columns \(X^B\) and \(Y\) (father's gametes? Wait, no, father is \(X^BY\)? Wait, no, the mother is \(X^BX^b\) (carrier) and father is \(X^BY\)? Wait, the Punnett square: left side (rows) are mother's gametes: \(X^B\) and \(X^b\), top (columns) are father's gametes: \(X^B\) and \(Y\). So the four offspring: \(X^BX^B\) (female, normal), \(X^BY\) (male, normal), \(X^bX^B\) (female, carrier), \(X^bY\) (male, color blind). So 1 out of 4 is color blind. So 25%? Wait, but wait, 1/4 is 25%, so the answer is 25 percent? Wait, no, wait: 1 out of 4 is 25%? Wait, 4 cells, 1 affected. So probability is 25%? Wait, but let's calculate again. Number of affected offspring: 1 (\(X^bY\)). Total offspring: 4. So \(\frac{1}{4}= 25\%\). Wait, but I think I made a mistake earlier. Wait, 1 out of 4 is 25%, so the probability is 25%.

Answer:

25 percent (Option: 25 percent)