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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

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

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### 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. 0 3. 1 4. \( \frac{1}{2} \) --- ### 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\(^H\)Y Where: - 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
Transcribed Image Text:### 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. 0 3. 1 4. \( \frac{1}{2} \) --- ### 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\(^H\)Y Where: - 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
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