Question 7 (1 point) An X-linked recessive gene causes cystic fibrosis in humans, a disorder affecting the function of the lungs . A woman with cystic fibrosis marries a normal man.. She is pregnant with their son. What is the probability that he will be born with cystic fibrosis? 0% 25% 50% 100%

The answer is 100%. Here's why:<br /><br />* **X-linked recessive inheritance:** The question states cystic fibrosis is X-linked recessive. This means the gene responsible is located on the X chromosome, and two copies of the recessive allele are needed for the condition to manifest in females (XX). Males (XY) only need one copy of the recessive allele on their single X chromosome to have the condition.<br /><br />* **Mother's genotype:** The woman has cystic fibrosis, meaning she must have two copies of the recessive allele (let's represent it as 'c'). Her genotype is XcXc.<br /><br />* **Father's genotype:** The man is normal. Since the gene is X-linked, his genotype must be XCY (where 'C' represents the dominant, normal allele).<br /><br />* **Possible offspring:** Let's create a Punnett square:<br /><br />```<br /> | XC | Y<br /> ----|-----|----<br /> Xc | XCXc| XcY<br /> Xc | XCXc| XcY<br />```<br /><br />* **Outcome:** As you can see, all possible male offspring (XcY) will inherit the recessive allele from their mother and will have cystic fibrosis. All possible female offspring (XCXc) will be carriers but will not have the condition.<br /><br />Therefore, the probability that their son will be born with cystic fibrosis is 100%. It's important to note that the question incorrectly states cystic fibrosis is X-linked. In reality, cystic fibrosis is caused by a mutation on chromosome 7 and follows an autosomal recessive inheritance pattern. However, based on the information *given in the question*, the logic and solution provided above are correct.<br />