NS FL3 B/B Q2

aeque
Posts: 3
Joined: Sat Jun 27, 2020 8:26 pm

NS FL3 B/B Q2

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!!!
NS_Tutor_Yuqi
Posts: 57
Joined: Fri May 29, 2020 11:43 pm

Re: NS FL3 B/B Q2

Hey! I read over the NS answer explanation for this and I feel like they went into way more detail than was actually necessary for solving the problem. You won't actually need to draw a punnet square in order to figure out the answer. We know from the question stem that disjunction occurs in 1/130 of live births. All instances of disjunction will lead to abnormal phenotypes. Therefore, in order to find the likelihood of the child having an normal phenotype, we will simply subtract 1/130 from 1. 1 - 0.008 - 0.992, or 99.2%. It sounds like you were confused about the assumption that disjunction automatically means that the mother passes down either no copies or two copies of the X chromosome and you used the punnet square in order to show that some children with disjunction will still have a normal phenotype. However, I think the error occurred when you introduced cases where the mother only passed down one X chromosome, since that is not actually a case of disjunction. Therefore, there aren't any instances of disjunction where the child will be phenotypically normal. I hope that helps answer your question a bit, but please let me know if I missed your point or if you want me to elaborate at all. Again, the answer explanation looks difficult but this is actually a relatively simple question.
aeque
Posts: 3
Joined: Sat Jun 27, 2020 8:26 pm

Re: NS FL3 B/B Q2

Thanks Yuqi,

Just to clarify this point:

"Disjunction automatically means that the mother passes down either no copies or two copies of the X chromosome"
So when considering disjunction questions, we should consider it as the gamete fate rather than the process? In my example, you can see that there are possible gametes after non-disjunction occurs in Meiosis 1 or Meiosis 2 there are potentially normal gametes. So even though we may say that the non-disjunction occurred during the process, if the gamete is a normal ploidy then we would say no non-disjunction occurred?

Or is it rather I should have been reading the question stem more closely. Since it reads "1/130 live births", this automatically means the fate of the gamete to be aneuploidy is 1/130. Rather if the question said the chance a "primary ova" goes through nondisjunction is 1/130, then my previous approach would have been correct?

Thanks!
NS_Tutor_Nancy
Posts: 48
Joined: Wed May 27, 2020 7:25 pm

Re: NS FL3 B/B Q2

Hey there!

Yes, you are exactly right. I think you got a little caught up in the weeds here because you were looking at every possible case scenario of nondisjunction, when the question stem simplifies it for you by saying nondisjunction occurs in 1/130 *live* births. It is very unlikely for an MCAT question to require your approach because of the time it would take, but yes theoretically if the question stem were to talk about the probability with regards to the ova rather than viable/live births, your approach would make sense.