MCAT Biology Question 28: Answer and Explanation

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Question: 28

13. An individual who is phenotypically female is found to have only one copy of a disease-carrying recessive allele on the X chromosome, yet she demonstrates all of the classic symptoms of the disease. Geneticists determine she has a genotype that likely arose from nondisjunction in one of her parents. What is the likely genotype of this individual?

  • A. 46,XX (46 chromosomes, with XX for sex chromosomes)
  • B. 46,XY
  • C. 45,X
  • D. 47,XXY

Correct Answer: C

Explanation:

Nondisjunction refers to the incorrect segregation of homologous chromosomes during anaphase I, or of sister chromatids during anaphase II. In either case, one daughter cell ends up with two copies of related genetic material, while the other receives zero. Immediately, this should eliminate choices (A) and (B), which show a normal complement of chromosomes (46). An individual who has only one recessive disease-carrying allele, and yet still expresses the disease, likely does not have a dominant allele for the given trait. This is seen in males, who are hemizygous for many X-linked genes, and can also be seen in women with Turner syndrome (45,X), who have only one X chromosome. Thus, choice (C) is the answer.

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