Date | November 2020 | Marks available | 1 | Reference code | 20N.1.HL.TZ0.13 |
Level | Higher level | Paper | Paper 1 | Time zone | TZ0 / no time zone |
Command term | Question number | 13 | Adapted from | N/A |
Question
Testing the chromosomes of a girl with Turner syndrome produced the following karyogram.
[Source: Turner’s syndrome karyotype 45,XO. This female lacks the second X chromosome present in the
normal karyotype. Symptoms include short stature, neck webbing, elbow deformity, widely spaced nipples with shield chest, primary amenorrhea, sexual infantilism and sterility. The ovaries are reduced to fibrous streaks. Also known as XO syndrome or ovarian short-stature syndrome. Credit: Wessex Reg. Genetics Centre. Attribution 4.0 International (CC BY 4.0).]
The condition can result from non-disjunction occurring in anaphase I of meiosis in an egg cell. Two cells result from the first division, one of which would lead to Turner syndrome. Which chromosomes will be in the other cell (polar body) at the end of meiosis I?
A. 44 autosomes and X
B. 44 autosomes and XX
C. 22 autosomes and X
D. 22 autosomes and XX
Markscheme
D
Examiners report
This was also a question where the answer could only be reached by careful deduction. It was necessary to understand that the chromosome number is halved in the first division of meiosis, so the two daughter cells produced will normally have 22 autosomes and one sex chromosome each. The karyogram shows that Turner’s syndrome is due to having only one X chromosome and no Y. One of the two gametes that fuse to produce a cell with this karyotype must have no sex chromosome. This is due to non-disjunction, where both sex chromosomes pass into one cell, rather than segregating. The answer to the question is therefore D.