Fabry disease follows X-linked recessive inheritance, so a phenotypically normal father (genotype XY, with normal X) and carrier mother (genotype XᶜX, where Xᶜ carries the disease allele) have a 50% chance of having a son with the disease. Among sons specifically, half inherit the mother’s diseased X chromosome (XᶜY, affected), while the other half get her normal X (XY, normal). Thus, the probability is 50%.

Punnett Square Analysis

The cross is: Father (XY) × Mother (XᶜX).

Each outcome has a 25% overall probability, but conditional on a son (50% of offspring), the affected son probability is 50%.

Option Explanations

No explicit options are listed, but common distractors for such questions include:

• 0%: Incorrect; assumes no transmission to sons (true for affected fathers, not carriers).

• 25%: Incorrect; confuses overall offspring probability with son-specific.

• 50%: Correct; son inherits mother’s X chromosome, 50% chance it’s diseased.

• 100%: Incorrect; would require homozygous mother.

Fabry disease X-linked inheritance makes it crucial for competitive exams like IIT JAM Biotechnology. This Fabry disease X-linked probability carrier mother son scenario tests core genetics: a phenotypically normal father (XY) and carrier mother (XᶜX) produce sons with 50% disease risk. Early mastery prevents errors in pedigree analysis.

Genetics of Fabry Disease

Fabry disease stems from GLA gene mutations on X chromosome, causing lysosomal storage issues. Males (XᶜY) show full symptoms; carrier females vary due to X-inactivation. Carrier mother passes diseased allele to 50% offspring regardless of sex.

Step-by-Step Probability Calculation

1. Mother contributes Xᶜ or X (50% each) to son; father always gives Y.

2. Son XᶜY: affected (50%); XY: normal (50%).

3. Answer: 50 (no decimal needed for integer fill-in).

Exam Tips for X-Linked Questions

• Focus on son’s X from mother only.

• Use Punnett squares for clarity.

• Differentiate carrier vs. affected parents.