### Probability Calculation of Hemophilic Offspring**Question:** What is the probability of a hemophilic daughter if the mother is a carrier and the father is normal? Remember, hemophilia is caused by a recessive X-linked allele.#### Answer Choices:1. \( \frac{3}{4} \)2. 03. 14. \( \frac{1}{2} \)---### Explanation:**Genetics Background:**Hemophilia is an X-linked recessive disorder, meaning it is carried on the X chromosome. Females have two X chromosomes (XX), while males have one X and one Y chromosome (XY). For a female to express hemophilia, she must inherit the hemophilia allele from both parents. However, she can be a carrier if she inherits the hemophilia allele from one parent.**Possible Genotypes:**- **Mother (Carrier):** X\(^H\)X\(^h\) - **Father (Normal):** X\(^H\)YWhere:- X\(^H\) represents the normal allele.- X\(^h\) represents the hemophilia allele.**Punnett Square Analysis:**To determine the probability of having a hemophilic daughter, we can create a Punnett Square combining the mother's X\(^H\)X\(^h\) and father's X\(^H\)Y genotypes.| | X\(^H\) (Father) | Y (Father) ||------|-----------------|------------|| X\(^H\) (Mother) | X\(^H\)X\(^H\) | X\(^H\)Y || X\(^h\) (Mother) | X\(^H\)X\(^h\) | X\(^h\)Y |From the Punnett Square, the possible offspring are:- X\(^H\)X\(^H\): Normal daughter- X\(^H\)X\(^h\): Carrier daughter- X\(^H\)Y: Normal son- X\(^h\)Y: Hemophilic son**Conclusion:**- There are no X\(^h\)X\(^h\) (hemophilic daughter) combinations in this scenario.- Hence, the probability of

A uncommon hereditary condition called haemophilia impairs the body's capacity to coagulate blood.…

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What is the probability of a hemophiliac daughter if the mother is a carrier and the father is normal? Remember hemophilia is caused by a recessive X-linked allele. 1/2 01 0 3/4 OC