Genetics Quiz: Nondisjunction

What this quiz covers

This quiz focuses on Nondisjunction, giving you a quick way to practice the rules, question types, and explanations that matter most for Genetics.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Anaphase lag produces only nullisomic and normal gametes, while nondisjunction produces only disomic and nullisomic gametes. (correct answer)
2. Anaphase lag produces only disomic and normal gametes, while nondisjunction produces only disomic and nullisomic gametes.
3. Both events produce the same portfolio of gametes, but anaphase lag is a result of spindle failure rather than chromatid cohesion.
4. Anaphase lag produces only monosomic zygotes upon fertilization, while nondisjunction can produce monosomic or trisomic zygotes.

Explanation: In Meiosis I nondisjunction, the homologous pair fails to separate, with both homologs moving to one pole. This leads to one secondary meiocyte being n+1 and the other being n-1, ultimately producing two disomic (n+1) and two nullisomic (n-1) gametes. In contrast, during anaphase lag in Meiosis I, one chromosome from a homologous pair is lost entirely. This results in one secondary meiocyte being normal (n) and the other being n-1. The subsequent Meiosis II division yields two normal (n) gametes and two nullisomic (n-1) gametes. Therefore, anaphase lag does not produce disomic (n+1) gametes, which is a key difference.

Question 2

1. A normal diploid cell line with one paternal and one maternal chromosome 13. (correct answer)
2. A diploid cell line exhibiting paternal uniparental isodisomy for chromosome 13.
3. A diploid cell line exhibiting maternal uniparental isodisomy for chromosome 13.
4. A diploid cell line exhibiting paternal uniparental heterodisomy for chromosome 13.

Explanation: A maternal Meiosis I error means the oocyte contained both of the mother's homologous chromosome 13s (M1 and M2). After fertilization by a normal paternal sperm (P1), the zygote is M1/M2/P1. Trisomy rescue involves the random loss of one of these three chromosomes. If the paternal chromosome (P1) is lost, the resulting cell is M1/M2, which is maternal uniparental heterodisomy. If one of the maternal chromosomes (e.g., M1) is lost, the resulting cell is M2/P1, which is a normal diploid cell line with biparental inheritance. Therefore, a normal cell line is a possible outcome. Isodisomy (two copies of the same homolog) would result from a Meiosis II error, not a Meiosis I error.

Question 3

1. Nondisjunction during maternal Meiosis I.
2. Nondisjunction during paternal Meiosis I. (correct answer)
3. Nondisjunction during maternal Meiosis II.
4. Nondisjunction during paternal Meiosis II.

Explanation: The child has three copies of chromosome 21. The presence of two different paternal alleles indicates that the child inherited both of the father's homologous chromosomes for this region. The failure of homologous chromosomes to separate occurs during Meiosis I. Therefore, the nondisjunction event happened during paternal Meiosis I, producing a sperm containing both of his homologous chromosome 21s. A Meiosis II error would result in the child receiving two copies of the same paternal allele, as it involves the failure of sister chromatids to separate.

Question 4

1. One cell with a 46,XX karyotype and one cell with a 47,XX,+21 karyotype.
2. One cell with a 47,XX,+21 karyotype and one cell with a 45,XX,-21 karyotype. (correct answer)
3. Both cells will have a 46,XX karyotype but will show mosaic expression.
4. Both cells will have a 47,XX,+21 karyotype, as the error is passed on.

Explanation: Mitotic nondisjunction is the failure of sister chromatids to separate during anaphase of mitosis. In the first division of a 46,XX zygote, the 46 chromosomes replicate to form 92 chromatids. If the two sister chromatids of chromosome 21 fail to separate, one daughter cell will receive both chromatids (resulting in a total of 47 chromosomes, i.e., Trisomy 21), while the other daughter cell will receive none (resulting in a total of 45 chromosomes, i.e., Monosomy 21). This establishes mosaicism from the earliest stage of development.

Question 5

1. Nondisjunction during paternal Meiosis I.
2. Nondisjunction during paternal Meiosis II.
3. Nondisjunction during maternal Meiosis I.
4. Nondisjunction during maternal Meiosis II. (correct answer)

Explanation: The woman's father was colorblind (X^cY), so she is an obligate carrier with genotype X^CX^c. The man has normal vision (X^CY). Their son is colorblind and has Klinefelter syndrome, so his genotype is X^cX^cY. He must have received the Y chromosome from his father. Therefore, he must have received an X^cX^c gamete from his mother. For a mother with genotype X^CX^c to produce an X^cX^c egg, nondisjunction must occur during Meiosis II. In Meiosis I, the homologous X^C and X^c chromosomes separate. The secondary oocyte destined to become the egg receives the X^c chromosome (which has replicated into two sister chromatids). Failure of these sister chromatids to separate in Meiosis II results in an X^cX^c egg. Nondisjunction in Meiosis I would have produced an X^CX^c egg.

Question 6

1. They would be identical sister chromatids.
2. They would be non-identical, homologous chromosomes. (correct answer)
3. They would be isodisomic.
4. They would exhibit segmental deletions due to the error.

Explanation: Nondisjunction in Meiosis I is a failure of homologous chromosomes to separate. Therefore, the resulting abnormal sperm would contain both of the father's homologous chromosomes 21 (one he inherited from his mother, one from his father). These two chromosomes are homologous but not identical, as they will have different alleles at many loci. This condition in the zygote is called heterodisomy. Sister chromatids are identical copies that fail to separate in Meiosis II nondisjunction, which results in isodisomy. Segmental deletions are a different type of chromosomal mutation.

Question 7

1. Nondisjunction in the purple-flowered parent, producing a nullisomic (n-1) gamete. (correct answer)
2. Nondisjunction in the white-flowered parent, producing a disomic (n+1) gamete.
3. A new point mutation in the P allele in the F1 zygote, changing it to p.
4. Nondisjunction in the purple-flowered parent, producing a disomic (n+1) gamete.

Explanation: The purple parent is PP and produces gametes with the P allele. The white parent is pp and produces gametes with the p allele. A normal F1 progeny would be Pp and have purple petals. The observed F1 plant has white petals, meaning it lacks the dominant P allele. Its genotype must be pO (where O is the missing chromosome 2). This plant is monosomic. To get this genotype, it must have inherited a gamete with the p allele from the white-flowered parent and a gamete lacking chromosome 2 (a nullisomic gamete) from the purple-flowered parent. The production of a nullisomic gamete is a consequence of nondisjunction.

Question 8

1. Autosomal aneuploidies are generally better tolerated and lead to more viable births than sex chromosome aneuploidies.
2. Sex chromosome nondisjunction primarily affects fertility, while autosomal nondisjunction leads to more severe congenital syndromes.
3. Nondisjunction occurs at a much higher frequency for sex chromosomes than for autosomes due to their size difference.
4. Dosage compensation mechanisms do not exist for autosomes, whereas X-inactivation mitigates the effects of extra X chromosomes. (correct answer)

Explanation: The key reason sex chromosome aneuploidies (like XXX, XXY, XYY) are more viable and have milder phenotypes than autosomal aneuploidies (like Trisomy 13, 18, 21) is the process of X-inactivation. This mechanism silences most of the genes on all but one X chromosome, balancing the gene dosage. Autosomes lack a comparable system-wide dosage compensation mechanism, so an extra autosome leads to a 150% expression of all its genes, which is often catastrophic for development. In fact, most autosomal trisomies are embryonic lethal. Therefore, choice D provides the most accurate underlying reason for the observed difference in severity.

Question 9

1. Two sperm of genotype (B,b) and two sperm nullisomic for chromosome 15.
2. One sperm of genotype (B,B), one sperm nullisomic for chromosome 15, and two sperm of genotype (b). (correct answer)
3. Two sperm of genotype (B) and two sperm of genotype (b).
4. One sperm of genotype (B,b), one nullisomic sperm, and two normal sperm of genotypes (B) and (b).

Explanation: In Meiosis I, homologous chromosomes separate. Nondisjunction here would lead to one secondary spermatocyte receiving both homologous chromosomes (containing B and b) and the other receiving none. This would result in two disomic (B,b) sperm and two nullisomic sperm. In contrast, Meiosis I proceeds normally in the setup for a Meiosis II error. One secondary spermatocyte receives the chromosome with the 'B' allele (replicated) and the other receives the chromosome with the 'b' allele (replicated). If nondisjunction of sister chromatids occurs in the 'B' cell, it produces one (B,B) sperm and one nullisomic sperm. The 'b' cell divides normally, producing two normal (b) sperm. Thus, this combination is unique to a Meiosis II error.

Question 10

1. Two sperm of genotype (B,b) and two sperm nullisomic for chromosome 15.
2. One sperm of genotype (B,B), one sperm nullisomic for chromosome 15, and two sperm of genotype (b). (correct answer)
3. Two sperm of genotype (B) and two sperm of genotype (b).
4. One sperm of genotype (B,b), one nullisomic sperm, and two normal sperm of genotypes (B) and (b).

Explanation: In Meiosis I, homologous chromosomes separate. Nondisjunction here would lead to one secondary spermatocyte receiving both homologous chromosomes (containing B and b) and the other receiving none. This would result in two disomic (B,b) sperm and two nullisomic sperm. In contrast, Meiosis I proceeds normally in the setup for a Meiosis II error. One secondary spermatocyte receives the chromosome with the 'B' allele (replicated) and the other receives the chromosome with the 'b' allele (replicated). If nondisjunction of sister chromatids occurs in the 'B' cell, it produces one (B,B) sperm and one nullisomic sperm. The 'b' cell divides normally, producing two normal (b) sperm. Thus, this combination is unique to a Meiosis II error.

Question 11

1. One cell with a 46,XX karyotype and one cell with a 47,XX,+21 karyotype.
2. One cell with a 47,XX,+21 karyotype and one cell with a 45,XX,-21 karyotype. (correct answer)
3. Both cells will have a 46,XX karyotype but will show mosaic expression.
4. Both cells will have a 47,XX,+21 karyotype, as the error is passed on.

Explanation: Mitotic nondisjunction is the failure of sister chromatids to separate during anaphase of mitosis. In the first division of a 46,XX zygote, the 46 chromosomes replicate to form 92 chromatids. If the two sister chromatids of chromosome 21 fail to separate, one daughter cell will receive both chromatids (resulting in a total of 47 chromosomes, i.e., Trisomy 21), while the other daughter cell will receive none (resulting in a total of 45 chromosomes, i.e., Monosomy 21). This establishes mosaicism from the earliest stage of development.

Question 12

1. Nondisjunction during maternal Meiosis I.
2. Nondisjunction during paternal Meiosis I. (correct answer)
3. Nondisjunction during maternal Meiosis II.
4. Nondisjunction during paternal Meiosis II.

Explanation: The child has three copies of chromosome 21. The presence of two different paternal alleles indicates that the child inherited both of the father's homologous chromosomes for this region. The failure of homologous chromosomes to separate occurs during Meiosis I. Therefore, the nondisjunction event happened during paternal Meiosis I, producing a sperm containing both of his homologous chromosome 21s. A Meiosis II error would result in the child receiving two copies of the same paternal allele, as it involves the failure of sister chromatids to separate.

Question 13

1. 2.0% (correct answer)
2. 3.0%
3. 4.0%
4. 6.0%

Explanation: This is a calculation of weighted averages.

1. Contribution from Meiosis I errors: 2% of meioses have this error. In these events, 50% of the resulting gametes are n+1. So, the contribution to the total gamete pool is 0.02 * 0.50 = 0.01 or 1%.
2. Contribution from Meiosis II errors: 4% of meioses have this error. In these events, 25% of the resulting gametes are n+1. So, the contribution to the total gamete pool is 0.04 * 0.25 = 0.01 or 1%.
3. Total frequency: The total frequency of n+1 gametes is the sum of the contributions from both types of errors: 1% + 1% = 2.0%. Distractor B (3.0%) comes from correctly calculating the M-I part (1%) and incorrectly calculating the M-II part as 4% * 0.5 = 2%. Distractor C (4.0%) comes from adding 2% and 4% but making calculation errors. Distractor D (6.0%) comes from adding the initial percentages (2%+4%).

Question 14

1. Increased rate of somatic mutations in the oocytes of older women.
2. Accumulation of environmental toxins that interfere with paternal spermatogenesis.
3. Degradation of sister chromatid cohesion proteins in oocytes arrested in Prophase I. (correct answer)
4. Higher frequency of paternal nondisjunction events in partners of older women.

Explanation: Human oocytes are arrested in Prophase I from fetal development until ovulation, which can be decades later. Cohesin proteins hold sister chromatids together. It is hypothesized that over this long arrest period, these cohesin complexes degrade or are damaged. This weakened cohesion increases the probability that homologous chromosomes or sister chromatids will fail to segregate correctly when meiosis resumes, leading to nondisjunction. Somatic mutations are changes in DNA sequence, not chromosome number. Paternal factors do not show a strong age-related correlation for nondisjunction compared to the maternal age effect.

Question 15

1. Autosomal aneuploidies are generally better tolerated and lead to more viable births than sex chromosome aneuploidies.
2. Sex chromosome nondisjunction primarily affects fertility, while autosomal nondisjunction leads to more severe congenital syndromes.
3. Nondisjunction occurs at a much higher frequency for sex chromosomes than for autosomes due to their size difference.
4. Dosage compensation mechanisms do not exist for autosomes, whereas X-inactivation mitigates the effects of extra X chromosomes. (correct answer)

Explanation: The key reason sex chromosome aneuploidies (like XXX, XXY, XYY) are more viable and have milder phenotypes than autosomal aneuploidies (like Trisomy 13, 18, 21) is the process of X-inactivation. This mechanism silences most of the genes on all but one X chromosome, balancing the gene dosage. Autosomes lack a comparable system-wide dosage compensation mechanism, so an extra autosome leads to a 150% expression of all its genes, which is often catastrophic for development. In fact, most autosomal trisomies are embryonic lethal. Therefore, choice D provides the most accurate underlying reason for the observed difference in severity.

Question 16

1. Nondisjunction in maternal meiosis, followed by fertilization by a normal X sperm.
2. Fertilization of a normal egg by a nullisomic sperm, followed by duplication of the X chromosome.
3. Formation of a normal 46,XX zygote, followed by loss of the maternal X chromosome during an early mitotic division. (correct answer)
4. Formation of a 47,XXX zygote, followed by the loss of two X chromosomes in one cell line.

Explanation: The individual has both a normal female cell line (46,XX) and a Turner syndrome cell line (45,X). This indicates the error occurred post-zygotically (mitotically). The zygote must have started as 46,XX. The problem states the remaining X in the 45,X line is paternal (Xp). This means that during a mitotic division of the original 46,XpXm zygote (p=paternal, m=maternal), the maternal X chromosome (Xm) was lost from one of the daughter cells due to anaphase lag or mitotic nondisjunction. This created the 45,Xp cell line, while the other lineage remained 46,XpXm.

Question 17

1. The child possesses two identical paternal alleles and one maternal allele.
2. The child possesses two different paternal alleles and one maternal allele. (correct answer)
3. The child is homozygous for a marker for which the father is also homozygous.
4. The child is heterozygous for a marker for which the mother is heterozygous and the father is homozygous.

Explanation: A paternal Meiosis II nondisjunction event is a failure of sister chromatids to separate. This would result in a sperm carrying two identical copies of a particular chromosome 18. Therefore, the resulting trisomic child would have two identical paternal alleles for any given marker (a condition known as uniparental isodisomy for the paternal contribution). The finding that the child possesses two different paternal alleles is indicative of uniparental heterodisomy, which results from a paternal Meiosis I nondisjunction event (failure of homologous chromosomes to separate). Therefore, this finding is inconsistent with a Meiosis II error.

Question 18

1. They would be identical sister chromatids.
2. They would be non-identical, homologous chromosomes. (correct answer)
3. They would be isodisomic.
4. They would exhibit segmental deletions due to the error.

Explanation: Nondisjunction in Meiosis I is a failure of homologous chromosomes to separate. Therefore, the resulting abnormal sperm would contain both of the father's homologous chromosomes 21 (one he inherited from his mother, one from his father). These two chromosomes are homologous but not identical, as they will have different alleles at many loci. This condition in the zygote is called heterodisomy. Sister chromatids are identical copies that fail to separate in Meiosis II nondisjunction, which results in isodisomy. Segmental deletions are a different type of chromosomal mutation.

Question 19

1. Increased rate of somatic mutations in the oocytes of older women.
2. Accumulation of environmental toxins that interfere with paternal spermatogenesis.
3. Degradation of sister chromatid cohesion proteins in oocytes arrested in Prophase I. (correct answer)
4. Higher frequency of paternal nondisjunction events in partners of older women.

Explanation: Human oocytes are arrested in Prophase I from fetal development until ovulation, which can be decades later. Cohesin proteins hold sister chromatids together. It is hypothesized that over this long arrest period, these cohesin complexes degrade or are damaged. This weakened cohesion increases the probability that homologous chromosomes or sister chromatids will fail to segregate correctly when meiosis resumes, leading to nondisjunction. Somatic mutations are changes in DNA sequence, not chromosome number. Paternal factors do not show a strong age-related correlation for nondisjunction compared to the maternal age effect.

Question 20

1. The child possesses two identical paternal alleles and one maternal allele.
2. The child possesses two different paternal alleles and one maternal allele. (correct answer)
3. The child is homozygous for a marker for which the father is also homozygous.
4. The child is heterozygous for a marker for which the mother is heterozygous and the father is homozygous.

Explanation: A paternal Meiosis II nondisjunction event is a failure of sister chromatids to separate. This would result in a sperm carrying two identical copies of a particular chromosome 18. Therefore, the resulting trisomic child would have two identical paternal alleles for any given marker (a condition known as uniparental isodisomy for the paternal contribution). The finding that the child possesses two different paternal alleles is indicative of uniparental heterodisomy, which results from a paternal Meiosis I nondisjunction event (failure of homologous chromosomes to separate). Therefore, this finding is inconsistent with a Meiosis II error.