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UNIT 05 / REVIEW

Nondisjunction and genetic evidence

Failure of homologs or sister chromatids to separate can produce aneuploid gametes. Use pedigrees and statistical tests as evidence, with explicit assumptions.

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01 / Nondisjunction and genetic evidence

Nondisjunction and genetic evidence

Meiosis I nondisjunction sends both homologs into one daughter cell; if meiosis II is otherwise normal, all four products are abnormal for that chromosome. Nondisjunction in one meiosis II cell can produce two normal and two abnormal products. A chi-square goodness-of-fit test compares observed counts with expected counts under a proposed model. A large deviation may reject that model, but it does not identify the biological cause by itself. Small pedigrees can be consistent with more than one inheritance model.

MAKE IT CONCRETE

For 80 offspring expected in a 3:1 ratio, expected counts are 60 and 20. Chi-square sums (observed − expected)^2 / expected for the two categories.

DON’T FALL FOR THIS

Failure to reject a hypothesis does not prove it true. Also, an extra copy of one chromosome is not an extra complete chromosome set.

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Original review notes informed by standard biology principles. Check the official AP Biology framework ↗ and OpenStax Biology 2e ↗ for further reading.