Difficulty: Intermediate | Prerequisites: meiosis stages (see companion study notes on Meiosis and Independent Assortment), basic understanding of dominance and recessiveness.
This set of notes covers the genetics of sex chromosomes: how X and Y chromosomes determine sex, how genes carried on the X chromosome produce distinctive inheritance patterns, and what happens when chromosomes fail to separate properly during meiosis (nondisjunction). It also covers pedigree analysis, the skill of reading a family tree to work out whether a trait is autosomal or sex-linked, dominant or recessive. You should already understand meiosis I and meiosis II, and be comfortable with terms like "allele," "dominant," and "recessive." If not, revisit the meiosis notes and your Chapter 4 material first.
Genes on the X chromosome follow different inheritance rules in males and females because males have only one X (they are hemizygous). Nondisjunction during meiosis produces individuals with abnormal sex chromosome counts like XXY or XYY, and knowing which division failed tells you which parent contributed the error. Pedigree analysis is pattern recognition: you rule out inheritance modes that cannot explain the family tree and identify the one that fits.
Pseudoautosomal region (PAR)
Short segments at the tips of the X and Y chromosomes (PAR1 and PAR2) where the two chromosomes share homologous sequences and can recombine during meiosis. In simple terms, these are the small zones where X and Y "act like" a normal homologous pair.
PAR dose
A count of how many PAR1 + PAR2 sets an individual carries. Each sex chromosome contributes one PAR1 and one PAR2, and together (PAR1 + PAR2) that equals one dose.
Hemizygous
Having only one copy of a gene, rather than two. Males are hemizygous for X-linked genes because they carry only one X chromosome. Think of it as: there is no second allele to mask a recessive one.
Sex-linked gene (X-linked gene)
A gene located on the X chromosome. Because males have one X and females have two, these genes produce different phenotypic ratios in males and females.
Nondisjunction
The failure of chromosomes (in meiosis I) or sister chromatids (in meiosis II) to separate properly during cell division. The result is gametes with too many or too few chromosomes.
Pedigree
A diagram showing the inheritance of a trait through multiple generations of a family. Squares represent males, circles represent females, and filled symbols indicate affected individuals.
Autosomal recessive
An inheritance pattern in which a trait appears only when an individual carries two copies of the recessive allele (one from each parent) on a non-sex chromosome. Carriers (heterozygotes) are unaffected.
X-linked recessive
An inheritance pattern in which the gene is on the X chromosome and the trait appears in all males who carry one copy of the recessive allele (because they have no second X to mask it) and in females only when homozygous recessive.
Carrier
An individual who is heterozygous for a recessive allele: they do not show the trait but can pass the allele to offspring. For X-linked recessive traits, carriers are typically females (X^A X^a).
What the PAR is
The pseudoautosomal regions (PAR1 at the tip of the short arms, PAR2 at the tip of the long arms) are the only segments of the X and Y chromosomes that recombine during male meiosis.
Genes in the PAR are inherited in an autosomal-like pattern because both X and Y carry copies.
Counting PAR doses
Each sex chromosome (whether X or Y) contributes one PAR1 and one PAR2.
The convention in this course: PAR1 + PAR2 from one chromosome = 1 dose.
A normal male (XY) has 2 doses (one from X, one from Y).
A normal female (XX) has 2 doses (one from each X).
Worked example: XXY individual
An XXY individual has three sex chromosomes.
Each contributes one dose: X (dose 1) + X (dose 2) + Y (dose 3) = 3 doses of the PAR regions.
The answer to sample question 3 is 3.
The classic white-eye cross
In Drosophila, the gene for eye colour is on the X chromosome. Red (W or w+) is dominant; white (w) is recessive.
Cross: white-eyed female (X^w X^w) x red-eyed male (X^W Y).
Expected F1: All females are X^W X^w (red-eyed, carriers). All males are X^w Y (white-eyed, because they got their single X from their white-eyed mother).
The exceptional (rare) progeny
Occasionally, at low frequency, the F1 includes white-eyed females and red-eyed males.
These "exceptional" offspring result from nondisjunction of the X chromosomes in the mother during meiosis.
Sex chromosome composition of exceptional white-eyed F1 females
A white-eyed F1 female must have received no functional dominant W allele from either parent.
The mother (X^w X^w) could produce an egg with two X chromosomes (X^w X^w) through nondisjunction. When fertilised by a Y-bearing sperm from the father, the result is X^w X^w Y.
This individual is X^w X^w Y: she is female (because she has two X chromosomes, which is sufficient for female development in Drosophila even with a Y present), and she is white-eyed because both her X chromosomes carry the w allele.
Alternatively, a non-disjunction egg with zero X chromosomes fertilised by the father's X^W sperm would give X^W O: a red-eyed male (because in Drosophila, X0 is male).
The answer to sample question 4: the white-eyed F1 females are X^w X^w Y (or written as XXY with both X chromosomes carrying w).
The question
In which cell division (meiosis I or meiosis II) and in which parent (male or female) would nondisjunction produce an XYY male?
Reasoning step by step
An XYY male has one X and two Y chromosomes.
The X must come from the mother (her eggs carry X). So the extra Y must come from the father.
In the father, meiosis I separates the X from the Y. If nondisjunction happened in meiosis I, the resulting gamete would be XY (both sex chromosomes together), not YY.
In meiosis II, sister chromatids of the Y chromosome separate. If nondisjunction happens here, one gamete gets two copies of Y (YY) and the other gets none.
A YY sperm fertilising a normal X egg produces XYY.
Answer
Nondisjunction in meiosis II, in the father.
Key logic to remember
Meiosis I nondisjunction in the father produces XY or O gametes (whole chromosomes fail to separate).
Meiosis II nondisjunction in the father produces XX, YY, or O gametes (sister chromatids fail to separate).
To get two copies of the same sex chromosome (YY), you need meiosis II failure of that chromosome.
How to determine the inheritance pattern
The quiz sample (question 6) shows a pedigree for a rare trait and asks you to identify the most likely mode of inheritance. Work through this checklist:
Does the trait skip generations? If affected children come from unaffected parents, the trait is likely recessive.
Are affected individuals predominantly male? If so, consider X-linked recessive.
Does an affected father ever have an unaffected daughter? If he does, the trait cannot be X-linked dominant (because a daughter must inherit his X).
Does an affected father pass the trait to all sons? If so, consider Y-linked. If not, Y-linked is ruled out.
Is every child of an affected mother also affected? If so, consider mitochondrial inheritance.
Applying the checklist to the sample pedigree
The trait is rare and appears to skip generations, suggesting recessive.
Affected individuals include both males and females across different generations, making it autosomal rather than sex-linked.
Unaffected parents (e.g., III-7 and III-8) produce affected sons (IV-5, IV-6), confirming both parents are carriers.
The most likely answer for question 6a: autosomal recessive.
Determining genotypes from a pedigree (autosomal recessive example)
Affected individuals are homozygous recessive: aa.
Unaffected individuals who have an affected child must be carriers: Aa.
Individual III-7 in the sample pedigree is an unaffected male who has affected sons. He must be a carrier.
Genotype of III-7: Aa.
If the trait were X-linked, you would write genotypes with the chromosome notation (e.g., X^A Y), but since this pedigree is autosomal recessive, plain Aa is correct.
Nondisjunction is the mechanism behind conditions like Klinefelter syndrome (XXY) and Turner syndrome (X0) in humans, which affect physical development and fertility. Pedigree analysis is the foundation of genetic counselling: when a family has a history of a genetic disorder, the counsellor traces the pattern through a pedigree to estimate the probability that future children will be affected.
Students often assume that nondisjunction in meiosis I and meiosis II produce the same result. They do not. Meiosis I nondisjunction sends both homologues to one cell; meiosis II nondisjunction sends both sister chromatids of one chromosome to one cell. The gamete contents are different.
A frequent error is thinking that X-linked recessive traits cannot appear in females. They can, but only when the female is homozygous recessive (X^a X^a) or, in Drosophila, when nondisjunction produces a female with two recessive X chromosomes and a Y.
Students sometimes confuse "carrier" with "affected." A carrier of an autosomal recessive trait (Aa) shows no symptoms. Only homozygous recessive (aa) individuals are affected.
In pedigree analysis, students often jump to X-linked recessive without checking whether affected females exist. If affected females appear (especially when their father is unaffected), autosomal recessive is more likely than X-linked recessive.
⚠️ PAR dose counting is a quick calculation question. Remember: each sex chromosome = 1 dose, regardless of whether it is an X or a Y.
⚠️ The Drosophila white-eye cross is a classic exam problem. Know the standard cross (X^w X^w x X^W Y) and be ready to explain exceptional progeny via nondisjunction.
⚠️ For nondisjunction questions, always ask: "Which parent contributed the extra (or missing) chromosome, and did it happen because homologues failed to separate (meiosis I) or sister chromatids failed to separate (meiosis II)?"
⚠️ Pedigree questions often ask for both the inheritance mode and a specific individual's genotype. Always determine the mode first, then assign genotypes consistent with that mode.
True or false: An XXY individual has 2 doses of the PAR regions. (False. 3 doses: one per sex chromosome.)
Fill in the blank: Males are ______ for X-linked genes because they have only one X chromosome. (hemizygous)
True or false: In Drosophila, sex is determined solely by the presence or absence of a Y chromosome. (False. Drosophila sex is determined by the X-to-autosome ratio. An XX individual is female even with a Y present.)
Fill in the blank: To produce an XYY male, nondisjunction must occur during meiosis ______ in the ______. (II; father)
True or false: If an autosomal recessive trait skips a generation, both parents in the skipped generation must be carriers. (True. Both must be Aa to produce an aa child.)
Q: In humans, how many doses of the PAR regions (PAR1 + PAR2 = 1 dose) are present in an individual with XXY sex chromosome constitution?
A: 3 doses. Each of the three sex chromosomes (X, X, and Y) contributes one dose.
Q: Red eye colour is X-linked dominant in Drosophila. A white-eyed female (X^w X^w) is crossed with a red-eyed male (X^W Y). What are the expected eye colours of the F1 females and males?
A: F1 females are X^W X^w (red-eyed, heterozygous carriers). F1 males are X^w Y (white-eyed, because they inherit their single X from their homozygous recessive mother).
Q: In the same cross, a low frequency of white-eyed females appear in the F1. What is their sex chromosome composition?
A: X^w X^w Y. These arise from nondisjunction in the mother, producing an egg with two X^w chromosomes that is then fertilised by a Y-bearing sperm.
Q: In which cell division and in which parent does nondisjunction give rise to an XYY male?
A: Meiosis II in the father. The Y chromosome's sister chromatids fail to separate, producing a YY sperm. This fertilises a normal X egg to give XYY.
Q: A rare trait appears in a pedigree where affected children come from unaffected parents, and both males and females are affected. What is the most likely inheritance pattern?
A: Autosomal recessive. The trait skipping generations (unaffected parents with affected children) indicates recessive inheritance, and both sexes being affected rules out X-linked as the most likely mode.
Q: In an autosomal recessive pedigree, an unaffected male has affected children. What is his genotype?
A: Aa (heterozygous carrier). He carries one copy of the recessive allele that he passes to affected offspring.
Nondisjunction ties back to the meiosis material in the companion study notes: it is the failure of the processes described there (anaphase I separation of homologues, anaphase II separation of sister chromatids). Understanding what normally happens in each stage is what lets you diagnose which stage went wrong.
Pedigree analysis connects forward to probability calculations in genetics: once you identify the inheritance pattern and carrier status, you can calculate the probability of an affected child using Mendelian ratios. It also connects to population genetics and Hardy-Weinberg equilibrium later in the course.
Sex chromosomes, X chromosome, Y chromosome, pseudoautosomal region, PAR1, PAR2, PAR dose, X-linked inheritance, X-linked recessive, sex-linked gene, hemizygous, Drosophila eye colour, white-eye gene, nondisjunction, meiosis I nondisjunction, meiosis II nondisjunction, XXY, XYY, Klinefelter, Turner, aneuploidy, pedigree analysis, autosomal recessive, autosomal dominant, carrier, genotype from pedigree, BIO 325 genetics, quiz 3 review