In a cross between two individuals with genotypes for haemophilia X^HX^h x X^HY, the chances of having a haemophilic son is

To determine the chances of having a haemophilic son, we need to perform a genetic cross using the given parental genotypes. Haemophilia is an X-linked recessive disorder. Step 1: Identify the genotypes of the parents. The problem states the genotypes are X^hX^h and X^HY. • Female parent: X^hX^h (haemophilic female) • Male parent: X^HY (normal male) Step 2: Set up a Punnett square for the cross. The female parent can produce only X^h gametes. The male parent can produce X^H and Y gametes. |c|c|c| Gametes & X^H & Y \\ X^h & X^HX^h & X^hY \\ X^h & X^HX^h & X^hY \\ Step 3: Determine the genotypes and phenotypes of the offspring. From the Punnett square, the possible offspring genotypes are: • X^HX^h: Carrier female (not haemophilic) • X^hY: Haemophilic male Step 4: Calculate the probability of having a haemophilic son. Out of 4 possible offspring combinations, 2 are X^hY (haemophilic sons). The probability of having a haemophilic son is (2)/(4) = (1)/(2). Step 5: Convert the probability to a percentage. (1)/(2) × 100\% = 50\% The chances of having a haemophilic son are 50%. The final answer is C. 50\%. Send me the next one 📸