Genetics Quiz: Meiosis Errors And Disorders

What this quiz covers

This quiz focuses on Meiosis Errors And Disorders, 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. Maternal Meiosis I (correct answer)
2. Maternal Meiosis II
3. Paternal Meiosis I
4. Paternal Meiosis II

Explanation: The child has alleles A, B, and C. The father must have contributed the C allele. Therefore, the mother must have contributed both the A and B alleles. Since the A and B alleles are on her two different homologous chromosomes 21, the error must have been a failure to separate these homologous chromosomes. This occurs during Meiosis I. If the error were in Meiosis II, she would have contributed two copies of the same allele (e.g., A/A or B/B).

Question 2

1. The spindle assembly checkpoint is unable to detect monosomy but can efficiently detect trisomy.
2. Genomic imprinting silences the single chromosome in a monosomy, resulting in a nullisomic state.
3. Autosomal monosomies arise from meiotic errors that are inherently more damaging than those causing trisomies.
4. Haploinsufficiency of numerous genes on the missing autosome is more detrimental than the gene dosage effect of a trisomy. (correct answer)

Explanation: When you encounter questions about chromosomal abnormalities and survival, focus on the fundamental principle of gene dosage balance. Cells are remarkably sensitive to having the correct amount of gene products, but they tolerate excess better than deficiency. The correct answer is D because haploinsufficiency explains why autosomal monosomies are lethal. When an entire autosome is missing, you lose one copy of hundreds or thousands of genes simultaneously. Many of these genes are haploinsufficient, meaning a single functional copy cannot produce enough protein for normal cellular function. This creates a catastrophic shortage of multiple essential gene products. In contrast, trisomies cause gene dosage imbalances through excess, which cells can often tolerate through various buffering mechanisms. Option A is incorrect because the spindle assembly checkpoint detects improper chromosome attachment regardless of whether the error will cause monosomy or trisomy. The checkpoint mechanism itself doesn't distinguish between these outcomes. Option B misrepresents genomic imprinting. While imprinting affects some genes, it doesn't silence entire chromosomes in monosomy cases. Most genes aren't imprinted and would still be expressed from the remaining chromosome. Option C incorrectly suggests that the meiotic errors themselves differ in severity. The same types of nondisjunction events can produce either monosomy or trisomy in different gametes. Remember this key principle: cells generally tolerate gene dosage excess better than deficiency. This explains why some trisomies are viable while autosomal monosomies are uniformly lethal—it's about haploinsufficiency, not the chromosomal error mechanism itself.

Question 3

1. Nondisjunction in paternal Meiosis I.
2. Nondisjunction in paternal Meiosis II. (correct answer)
3. Nondisjunction in maternal Meiosis I.
4. A post-zygotic mitotic error in the embryo.

Explanation: The child has paternal uniparental isodisomy, meaning both chromosomes are identical copies from the father. The child's genotype is 'aa', so both inherited chromosomes carry the 'a' allele. This requires a sperm that contained two copies of the chromatid carrying the 'a' allele. This occurs if paternal Meiosis I proceeds normally (separating the 'A' and 'a' homologs), but in Meiosis II, the sister chromatids of the chromosome carrying 'a' fail to separate. This produces an 'aa' sperm, which then fertilizes a nullisomic ('O') egg. Paternal Meiosis I nondisjunction would result in heterodisomy, and the child's genotype would be 'Aa'.

Question 4

1. Nondisjunction in paternal Meiosis I.
2. Nondisjunction in paternal Meiosis II. (correct answer)
3. Nondisjunction in maternal Meiosis I.
4. A post-zygotic mitotic error in the embryo.

Explanation: The child has paternal uniparental isodisomy, meaning both chromosomes are identical copies from the father. The child's genotype is 'aa', so both inherited chromosomes carry the 'a' allele. This requires a sperm that contained two copies of the chromatid carrying the 'a' allele. This occurs if paternal Meiosis I proceeds normally (separating the 'A' and 'a' homologs), but in Meiosis II, the sister chromatids of the chromosome carrying 'a' fail to separate. This produces an 'aa' sperm, which then fertilizes a nullisomic ('O') egg. Paternal Meiosis I nondisjunction would result in heterodisomy, and the child's genotype would be 'Aa'.

Question 5

1. The original zygote was 46,XX and a mitotic nondisjunction event created the trisomic cell line.
2. The original zygote was 47,XX,+13 and a mitotic anaphase lag led to loss of a chromosome 13. (correct answer)
3. The amniocentesis result was due to confined placental mosaicism and did not reflect the fetus.
4. The fetus inherited a balanced translocation from one parent that became unstable after birth.

Explanation: The most likely sequence of events is that the zygote was constitutionally trisomic for chromosome 13, as indicated by the initial amniocentesis. Subsequently, during mitotic divisions in the developing embryo, a process called anaphase lag or mitotic nondisjunction resulted in the loss of one chromosome 13 in a cell. This event, known as trisomy rescue, created a second, diploid (46,XX) cell line, leading to the observed mosaicism. While confined placental mosaicism is possible, finding mosaicism in the baby's blood makes trisomy rescue in the embryo a more direct explanation.

Question 6

1. Nondisjunction of chromosome 5 during meiosis II.
2. Unequal crossing over between misaligned repetitive DNA sequences. (correct answer)
3. Formation of an isochromosome of the long arm of chromosome 5.
4. A Robertsonian translocation involving chromosome 5.

Explanation: While spontaneous chromosome breakage can cause terminal deletions, a common mechanism for recurrent microdeletions is unequal crossing over during meiosis. If homologous chromosomes misalign at regions with repetitive DNA sequences (like low-copy repeats) and a crossover occurs, the resulting recombinant chromatids can have a duplication or a deletion of the segment between the misaligned repeats. This is a well-established mechanism for many de novo microdeletion syndromes. Nondisjunction causes aneuploidy (wrong number), isochromosome formation involves centromere misdivision, and Robertsonian translocations only involve acrocentric chromosomes (13, 14, 15, 21, 22).

Question 7

1. Adjacent-1 segregation
2. Adjacent-2 segregation
3. Alternate segregation
4. 3:1 segregation (correct answer)

Explanation: When you encounter chromosomal translocation problems, focus on understanding how different segregation patterns during meiosis lead to specific karyotypes in offspring. In this case, the mother carries a balanced t(11;22) translocation, meaning she has four chromosomes that can pair during meiosis: normal 11, normal 22, derivative 11, and derivative 22. The child's karyotype shows 47 chromosomes with an extra derivative 22, indicating he received both a normal chromosome 22 and the derivative 22 from his mother, plus a normal 22 from his father. This pattern results from 3:1 segregation (choice D), where three chromosomes go to one gamete and one goes to the other. Specifically, the mother's gamete contained both the normal 22 and derivative 22, creating the unbalanced situation that leads to Emanuel syndrome when combined with the father's normal contribution. Adjacent-1 segregation (A) would produce gametes with either both derivatives or both normal chromosomes, not the mix seen here. Adjacent-2 segregation (B) involves homologous chromosomes going to the same gamete, which doesn't match this karyotype pattern. Alternate segregation (C) produces balanced gametes with either normal chromosomes or both derivatives, resulting in either normal offspring or balanced carriers. Study tip: For translocation problems, always count the total chromosome number first, then trace which specific chromosomes are present. An abnormal total (like 47) immediately points you toward 3:1 segregation, while normal totals (46) suggest adjacent or alternate patterns.

Question 8

1. Lack of recombination prevents the proper degradation of securin, leading to premature sister chromatid separation.
2. Chiasmata formed by crossovers are required to physically tether homologous chromosomes for proper spindle attachment. (correct answer)
3. The absence of crossover events leads to hyper-compaction of chromosomes, which physically obstructs segregation.
4. Recombination is necessary to activate the spindle assembly checkpoint; without it, errors go undetected.

Explanation: Crossing over results in the formation of chiasmata, which are physical links that hold homologous chromosomes together after the synaptonemal complex dissolves. These chiasmata create tension when microtubules from opposite spindle poles attach to the homologous kinetochores. This tension is crucial for stabilizing the attachment and ensuring proper alignment on the metaphase I plate. Without at least one chiasma, the homologous chromosomes (a bivalent) can behave as univalents, attaching to the spindle improperly and leading to their nondisjunction.

Question 9

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

Explanation: The male is XXY and has both G6PD-A and G6PD-B alleles. His mother is homozygous for G6PD-A, so she must have contributed an X chromosome with the G6PD-A allele. His father has the G6PD-B allele on his X chromosome. To have both alleles, the son must have received X(A) from his mother and an XY sperm from his father. An XY sperm is formed when the X and Y chromosomes fail to segregate during paternal Meiosis I.

Question 10

1. Fertilization of a normal oocyte by a diploid sperm from a paternal Meiosis I error.
2. Fertilization of a diploid oocyte from a maternal Meiosis I error by a Y-sperm.
3. Fertilization of a normal oocyte by two separate sperm (dispermy). (correct answer)
4. Failure of the first mitotic cleavage in a normal 46,XY zygote.

Explanation: Triploidy (69 chromosomes) can result from a diploid gamete (either egg or sperm) or from dispermy (fertilization of one egg by two sperm). Large-scale studies have shown that dispermy is the most common cause of human triploidy, accounting for approximately 60% of cases. Fertilization by a diploid sperm and fertilization of a diploid egg account for the remaining cases. Failure of the first mitotic cleavage would result in tetraploidy (92 chromosomes), not triploidy.

Question 11

1. An unbalanced gamete from adjacent-1 segregation, leading to partial trisomy and partial monosomy. (correct answer)
2. A high probability of producing only offspring with full trisomy 3 or full monosomy 11.
3. The formation of a quadrivalent structure during meiosis II that leads to obligatory chromosome loss.
4. An increased risk of mosaicism in the offspring due to instability of the translocated chromosomes.

Explanation: During Meiosis I, the translocated chromosomes and their normal homologs form a quadrivalent structure. Segregation can occur in several ways. Alternate segregation produces normal or balanced gametes. However, adjacent-1 and adjacent-2 segregation patterns lead to unbalanced gametes, which contain a duplication of one translocated segment and a deletion of the other. This partial aneuploidy is the most significant risk for producing a child with congenital abnormalities. Full aneuploidies are not the typical result, and the quadrivalent forms in Meiosis I, not II.

Question 12

1. Lack of recombination prevents the proper degradation of securin, leading to premature sister chromatid separation.
2. Chiasmata formed by crossovers are required to physically tether homologous chromosomes for proper spindle attachment. (correct answer)
3. The absence of crossover events leads to hyper-compaction of chromosomes, which physically obstructs segregation.
4. Recombination is necessary to activate the spindle assembly checkpoint; without it, errors go undetected.

Explanation: Crossing over results in the formation of chiasmata, which are physical links that hold homologous chromosomes together after the synaptonemal complex dissolves. These chiasmata create tension when microtubules from opposite spindle poles attach to the homologous kinetochores. This tension is crucial for stabilizing the attachment and ensuring proper alignment on the metaphase I plate. Without at least one chiasma, the homologous chromosomes (a bivalent) can behave as univalents, attaching to the spindle improperly and leading to their nondisjunction.

Question 13

1. Failure of homologous chromosomes to separate during anaphase I.
2. A reciprocal translocation between two homologous chromosomes.
3. Fusion of the ends of a chromosome after loss of telomeres.
4. Transverse division of the centromere during anaphase II. (correct answer)

Explanation: When you encounter questions about chromosomal abnormalities, focus on understanding the specific mechanisms that create each type of structural defect. An isochromosome is a unique abnormality where both arms of a chromosome are identical, creating a structure that looks like two q arms or two p arms joined at the centromere. The correct answer is D because isochromosomes form when the centromere divides transversely (horizontally) instead of longitudinally (vertically) during anaphase II. Normally, sister chromatids separate when the centromere splits along its length. However, when division occurs across the centromere's width, you get two abnormal chromosomes: one with both p arms and one with both q arms. The i(Xq) in Turner syndrome represents the chromosome containing two identical long arms of the X chromosome. Option A describes nondisjunction during meiosis I, which causes entire homologous chromosomes to move to the same cell, resulting in aneuploidy, not structural chromosome abnormalities. Option B refers to reciprocal translocations, where chromosome segments are exchanged between non-homologous chromosomes, creating derivative chromosomes but not isochromosomes. Option C describes telomere fusion leading to dicentric chromosomes or ring chromosomes, not the symmetric structure of an isochromosome. Remember that isochromosome formation specifically involves abnormal centromere division orientation. When you see "identical arms" in a question, think about how the centromere must have divided incorrectly to create this mirror-image structure.

Question 14

1. The ring structure prevents gene transcription, effectively silencing the entire chromosome.
2. The ring chromosome is mitotically unstable, leading to somatic mosaicism with cells that have lost the ring. (correct answer)
3. The formation of the ring requires no loss of genetic material, so the phenotype is caused by altered topology.
4. Ring chromosomes replicate twice as fast as linear chromosomes, causing a functional trisomy.

Explanation: Ring chromosomes are notoriously unstable during mitosis. They can be lost entirely, or they can form interlocking rings or larger structures that are also lost or mis-segregated. This leads to somatic mosaicism, where an individual has a mixture of cell lines (e.g., some with the ring, some monosomic for chromosome 14). The proportion and distribution of these different cell lines throughout the body contribute significantly to the variability in clinical severity observed among patients.

Question 15

1. Accumulation of point mutations in the DNA of aging oocytes.
2. Decreased efficiency of the spindle assembly checkpoint in older women.
3. Progressive degradation of cohesin complexes holding homologous chromosomes together. (correct answer)
4. Increased frequency of de novo chromosome translocation events in the maternal germline.

Explanation: Human oocytes are arrested in prophase I from before birth until ovulation, a period that can last for decades. During this long arrest, the cohesin proteins that physically link homologous chromosomes (at chiasmata) and sister chromatids can degrade. This weakening of cohesion increases the likelihood that homologous chromosomes will separate prematurely or fail to orient correctly on the metaphase I spindle, leading to nondisjunction. While checkpoint efficiency may also decline, the primary physical cause is the loss of cohesion.

Question 16

1. Two normal chromatids and two chromatids with the balanced inversion.
2. Four chromatids, each with a different combination of terminal duplications and deletions.
3. One normal chromatid, one inverted chromatid, one dicentric chromatid, and one acentric fragment. (correct answer)
4. One normal chromatid, one inverted chromatid, and two recombinant chromatids that are chromosomally balanced.

Explanation: A crossover within the inversion loop of a paracentric inversion (where the centromere is outside the inverted segment) connects the two homologous chromosomes. When they are pulled apart in Anaphase I, this creates a dicentric chromatid (with two centromeres) that forms a bridge and an acentric fragment (with no centromere) that will be lost. The two non-recombinant chromatids, one normal and one inverted, are also produced. The dicentric bridge will break, and the resulting gametes will be unbalanced and likely non-viable.

Question 17

1. Nondisjunction of the X and Y chromosomes during Meiosis I. (correct answer)
2. Nondisjunction of sister chromatids of the X chromosome during Meiosis II.
3. Nondisjunction of sister chromatids of the Y chromosome during Meiosis II.
4. Formation of an isochromosome of the Y chromosome.

Explanation: In Meiosis I of spermatogenesis, the X and Y chromosomes pair and then segregate. If nondisjunction occurs, one secondary spermatocyte receives both the X and Y chromosome, while the other receives no sex chromosome ('O'). The XY secondary spermatocyte will then produce two XY sperm in Meiosis II. The 'O' secondary spermatocyte will produce two 'O' sperm. Therefore, nondisjunction in Meiosis I is the only single event that produces both XY and 'O' sperm from the same primary spermatocyte.

Question 18

1. Nondisjunction during maternal meiosis I. (correct answer)
2. Nondisjunction during maternal meiosis II.
3. Nondisjunction during paternal meiosis I.
4. A post-zygotic mitotic nondisjunction event in the embryo.

Explanation: The child inherited two distinct homologous chromosomes from the mother. Homologous chromosomes separate during meiosis I. Therefore, a failure of this separation (nondisjunction in meiosis I) would result in a gamete containing both homologous chromosomes. Nondisjunction in meiosis II involves the failure of sister chromatids to separate, which would result in a gamete with two identical copies of a single parental homologue. Paternal nondisjunction would result in the child inheriting two chromosomes from the father. A post-zygotic mitotic error would lead to mosaicism, which is not described.

Question 19

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

Explanation: The male is XXY and has both G6PD-A and G6PD-B alleles. His mother is homozygous for G6PD-A, so she must have contributed an X chromosome with the G6PD-A allele. His father has the G6PD-B allele on his X chromosome. To have both alleles, the son must have received X(A) from his mother and an XY sperm from his father. An XY sperm is formed when the X and Y chromosomes fail to segregate during paternal Meiosis I.

Question 20

1. An unbalanced gamete from adjacent-1 segregation, leading to partial trisomy and partial monosomy. (correct answer)
2. A high probability of producing only offspring with full trisomy 3 or full monosomy 11.
3. The formation of a quadrivalent structure during meiosis II that leads to obligatory chromosome loss.
4. An increased risk of mosaicism in the offspring due to instability of the translocated chromosomes.

Explanation: During Meiosis I, the translocated chromosomes and their normal homologs form a quadrivalent structure. Segregation can occur in several ways. Alternate segregation produces normal or balanced gametes. However, adjacent-1 and adjacent-2 segregation patterns lead to unbalanced gametes, which contain a duplication of one translocated segment and a deletion of the other. This partial aneuploidy is the most significant risk for producing a child with congenital abnormalities. Full aneuploidies are not the typical result, and the quadrivalent forms in Meiosis I, not II.