1. **Stating the problem:** We have a genetic inheritance problem involving a trait denoted by $x^{CB}$ and $x^{cb}$ on the X chromosome, with males and females having different genotypes. We want to find the percentages of affected and normal children by gender. 2. **Understanding the genetics:** - Males have one X chromosome (XY), females have two (XX). - $x^{CB}$ likely represents the affected allele, $x^{cb}$ the normal or carrier allele. - The problem asks about affected males, normal females, carrier females, and normal children percentages. 3. **Assuming the parents' genotypes:** - Mother: $x^{CB}x^{cb}$ (carrier female) - Father: $x^{cb}Y$ (normal male) 4. **Possible children genotypes:** - Sons (XY): inherit Y from father, X from mother - $x^{CB}Y$ (affected son) - $x^{cb}Y$ (normal son) - Daughters (XX): inherit X from both parents - $x^{CB}x^{cb}$ (carrier daughter) - $x^{cb}x^{cb}$ (normal daughter) 5. **Calculating probabilities:** - Sons: - Probability affected = $\frac{1}{2}$ (inherit $x^{CB}$ from mother) - Probability normal = $\frac{1}{2}$ (inherit $x^{cb}$ from mother) - Daughters: - Probability carrier = $\frac{1}{2}$ (inherit $x^{CB}$ from mother) - Probability normal = $\frac{1}{2}$ (inherit $x^{cb}$ from mother) 6. **Answering questions:** 1. Are there affected children? Yes, affected sons exist. 2. Percentage of affected sons = $50\%$ 3. Percentage of normal daughters = $50\%$ 4. Percentage of carrier daughters = $50\%$ 5. Percentage of normal children overall: - Total children: 50\% sons + 50\% daughters - Normal sons = 25\% (half of sons) - Normal daughters = 25\% (half of daughters) - Total normal children = $25\% + 25\% = 50\%$ Final answers: - Affected sons: $50\%$ - Normal daughters: $50\%$ - Carrier daughters: $50\%$ - Normal children overall: $50\%$