Calculating X-linked Recessive Disease Probability in Son
This question concerns the probability of inheriting an X-linked recessive disease. We are given that the mother is a carrier and the father is normal. We need to find the probability that their son is born with the disease.
Understanding Inheritance Patterns
- X-linked recessive inheritance means the disease-causing allele is located on the X chromosome, and it only causes the disease when an individual has two copies of the allele (in females) or one copy (in males, as they only have one X chromosome).
- Carrier Mother: A carrier female has one normal allele (let's denote it as XR) and one disease allele (Xr). Her genotype is XRXr.
- Normal Father: A normal male has one normal X chromosome and one Y chromosome. His genotype is XRY.
- Son's Inheritance: A son inherits his Y chromosome from his father and his X chromosome from his mother.
Probability Calculation
To determine if the son has the disease, we look at the X chromosome he inherits from his mother. The mother has two possible alleles to pass on: XR or Xr.
- The probability of the mother passing the normal allele (XR) is 1/2. If the son inherits this, his genotype is XRY, and he is normal.
- The probability of the mother passing the disease allele (Xr) is 1/2. If the son inherits this, his genotype is XrY. Since the disease is X-linked recessive, this genotype results in the son having the disease.
Therefore, the probability that the son is born with the disease is the probability of inheriting the Xr allele from the mother, which is:
$ P(\text{Son is affected}) = P(\text{Mother passes } X^r) = \frac{1}{2} = 0.5 $
Final Answer Check
The calculated probability is 0.5. Rounded off to one decimal place, this remains 0.5.
Important Questions from Mendelian Inheritance
- Mendel's 'law of segregation' applies to the segregation of __________ during gamete formation.
- Fabry disease in humans is a X-linked disease. The probability (in percentage) for a phenotypically normal father and a carrier mother to have a son with Fabry disease is ________.
The allele associated with albinism in humans is recessive ($c$). The probability that an albino male ($cc$) and a carrier female ($Cc$) will have an offspring with normal skin pigmentation is _________.
(Round off to one decimal place)
- The blood group of the mother is A⁺ and that of the father is AB⁺. Which of the following statements is/are correct?
- Assuming independent assortment and no recombination, the number of different combinations of maternal and paternal chromosomes in gametes of an organism with a diploid number of 12 is ________.
