### NS FL3 B/B Q2

Posted:

**Sat Jun 27, 2020 8:32 pm**Need help understanding this question:

"The instance of nondisjunction for the X chromosome in females over the age of 30 is about one out of every 130 live births. If a woman over 30 gives birth to a viable baby, assuming the risk of nondisjunction from the father is negligible, what is the likelihood that it will have a normal phenotype?"

a: 99.2% b: 99.6% c: 0.6% d: 0.4%

Obvious C and D are out because 1/130 is 0.8%. The answer next step provided was the answer is A because after non-disjunction there is a 100% chance of a x-linked mutation the mother will pass either XX or XO to her child. That doesn't seem right to me?

Left 4 are Mother with nondisjunction Meiosis I and the right 4 are from nondisjunction in meiosis 2; Dad's sperm is on the left hand side column. * denotes healthy viable genotype, ^ denotes nonviable children

XX XX O O X X XX O

X: XXX XXX XO XO XX* XX* XXX XO

Y: XXY XXY YO^ YO^ YY* XY* XXY YO^

So of the 16 possible combinations here, Only 13 are viable, and of that 4 are healthy. So the chance of a non-healthy but viable baby from non-disjunction should be 9/13 or ~70%.

70% * 0.8% = 0.06% thus the chance of a healthy baby is 99.4%.

When I took this I just said it has to be greater than 0.08% since there are possible viable outcomes from nondisjunction so I went with B (99.6%). Anyone have any thoughts here? I really doubt the real MCAT is gonna get this tricky on this but just want to make sure I truly understand this.

Thanks!!!

"The instance of nondisjunction for the X chromosome in females over the age of 30 is about one out of every 130 live births. If a woman over 30 gives birth to a viable baby, assuming the risk of nondisjunction from the father is negligible, what is the likelihood that it will have a normal phenotype?"

a: 99.2% b: 99.6% c: 0.6% d: 0.4%

Obvious C and D are out because 1/130 is 0.8%. The answer next step provided was the answer is A because after non-disjunction there is a 100% chance of a x-linked mutation the mother will pass either XX or XO to her child. That doesn't seem right to me?

Left 4 are Mother with nondisjunction Meiosis I and the right 4 are from nondisjunction in meiosis 2; Dad's sperm is on the left hand side column. * denotes healthy viable genotype, ^ denotes nonviable children

XX XX O O X X XX O

X: XXX XXX XO XO XX* XX* XXX XO

Y: XXY XXY YO^ YO^ YY* XY* XXY YO^

So of the 16 possible combinations here, Only 13 are viable, and of that 4 are healthy. So the chance of a non-healthy but viable baby from non-disjunction should be 9/13 or ~70%.

70% * 0.8% = 0.06% thus the chance of a healthy baby is 99.4%.

When I took this I just said it has to be greater than 0.08% since there are possible viable outcomes from nondisjunction so I went with B (99.6%). Anyone have any thoughts here? I really doubt the real MCAT is gonna get this tricky on this but just want to make sure I truly understand this.

Thanks!!!