## FL #3 B/B Question 2

heatfan
Posts: 25
Joined: Sun Apr 30, 2017 12:34 pm

### FL #3 B/B Question 2

Hello,

I wanted to know are they saying nondisjunction is 100 % or 1 because all these trisomies are a result of nondisjunction? Also, how are we supposed to calculate 1/130 when taking the exam? We are not given any calculator and it seems to be a strenuous calculations. Thanks.
NS_Tutor_Katelyn
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### Re: FL #3 B/B Question 2

heatfan wrote:Hello,

I wanted to know are they saying nondisjunction is 100 % or 1 because all these trisomies are a result of nondisjunction? Also, how are we supposed to calculate 1/130 when taking the exam? We are not given any calculator and it seems to be a strenuous calculations. Thanks.

If non-disjunction occurs, then there is a 100% chance of an abnormal phenotype. We know from the question that there is a 1/130 chance of non-disjunction occurring, which is a little less than 1%. Therefore, the odds of a phenotypically normal offspring would be about 99%. Of all the options, answer choice A is closest to that.

Hopefully that helps!

-Kate
Katelyn Sawyer
Senior Tutor
katelyn@nextsteptestprep.com