Triple X syndrome (also known as Trisomy X) results from a random nondisjunction

Question

Triple X syndrome (also known as Trisomy X) results from a random nondisjunction event either in the formation of parental gametes or during development of an embryo. This syndrome have variable expressivity, and for this problem, assume that it is 100% penetrant in the population. There are about 7 girls born with XXX syndrome each day (in the USA), and 1 in every 1000 girls are born with XXX syndrome. (Because of variable expressivity, some women don't show clear symptoms and don't even know they are formally XXX.) Given this information, what is the probability that a normal father and an asymptomatic XXX mother will give birth to a baby girl with only 1 Barr Body per somatic cell in her body? (Assume there is no mosaicism going on.) Enter your answer to 4 decimal places.

Classic Klinefelter Syndrome (KS) is characterized by males having an abnormal extra X chromosomes (47,XXY). Karyotyping is the definitive way of diagnosing KS given variable expressivity, although traits associated with KS is essentially 100% penetrant with little to no mosaicism. Only 10% of KS cases are found by prenatal testing, 3 for every 2000 males have KS, and about 3% of infertile males actually have KS. Given this and assuming that there are 10,830 babies born in the USA each day and 105 boys born to every 100 girls born, how many babies with KS having 1 Barr Body per somatic cell (assume no other sex chromosome anomalies in the population) are born each year in the USA that were not diagnosed at the prenatal testing stage as having KS? Enter a whole number (round up).

Explanation / Answer

For XXX syndrome,

a normal male will produce two types of sperms i.e. X and Y

asymptomatic female will produce two types of eggs i.e XX and X

Probability of having such mother = 1/1000 = 0.001

progeny of normal male and such mother will be

XXX and XX daughters

XXY and XY sons

Probability of girl with one barr body i.e. normal daughter is 1/4 i.e. 0.25

Probability of such girl including probability of having mother with XXX syndrome = 0.25 × 0.001 = 0.00025 or 0.0003

For Klinefelter's syndrome

In USA, probability of having boy = 105/100+105 = 105/205

Probability of boy having KS = 3/2000

probability of getting KS undiagnosed = 1 - (10/100) = 1- 0.1 = 0.9

Total probability = (105/205) × (3/2000) × 0.9 = 0.00069

Total babies with KS = 0.00069 × 10830 = 7.48 or 8 babies

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