- A zygote is conceived with trisomy 16, a condition that is typically embryonic lethal. In an early blastomere, a mitotic nondisjunction event leads to the loss of one chromosome 16, a process known as trisomy rescue. Which of the following is a potential consequence for the developing fetus?
Answer: The resulting diploid cell line will exhibit paternal uniparental disomy if the maternal chromosome is lost.
Trisomy rescue can lead to a diploid cell line, but it also creates mosaicism (a mix of trisomic and diploid cells). Of the three chromosomes 16 present, one is paternal and two are maternal (assuming maternal nondisjunction, the most common cause). If the rescue event randomly eliminates one of these three, there is a 1/3 chance of losing the paternal chromosome, resulting in a diploid cell with two maternal chromosomes (maternal uniparental disomy). There is also a 2/3 chance of losing one of the maternal chromosomes, resulting in a normal diploid cell line. Therefore, uniparental disomy is a significant potential outcome.
- A phenotypically normal woman carries a balanced reciprocal translocation, 46,XX,t(3;11)(p12;q23). Which of the following represents the most significant risk for a viable pregnancy if her partner has a normal karyotype?
Answer: An unbalanced gamete from adjacent-1 segregation, leading to partial trisomy and partial monosomy.
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.
- While several autosomal trisomies and sex chromosome aneuploidies are compatible with postnatal life, autosomal monosomies are almost uniformly embryonic lethal. What is the primary genetic principle that explains this observation?
Answer: Haploinsufficiency of numerous genes on the missing autosome is more detrimental than the gene dosage effect of a trisomy.
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.
- A child with Down syndrome (Trisomy 21) is genotyped for a highly polymorphic microsatellite marker near the centromere of chromosome 21. The child's genotype is A/B/C. The mother's genotype is A/B, and the father's genotype is C/D. In which parent and at what stage did the nondisjunction event occur?
Answer: Maternal Meiosis I
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).
- A child is diagnosed with a recessive disorder (genotype 'aa'). The mother's genotype is 'AA' (homozygous wild-type), and the father's is 'Aa' (heterozygous). Cytogenetic analysis reveals the child has a normal chromosome number but has paternal uniparental isodisomy for the chromosome carrying this gene. What specific meiotic error is consistent with these findings?
Answer: Nondisjunction in paternal Meiosis II.
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'.
- A newborn presents with features of Emanuel syndrome. The karyotype report is 47,XY,+der(22)t(11;22)(q23;q11.2)mat. The mother is a known carrier of the balanced t(11;22) translocation. This specific unbalanced karyotype in the child is the result of what meiotic segregation pattern in the mother?
Answer: 3:1 segregation
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.
- A child is diagnosed with Trisomy 18. Genetic analysis using polymorphic markers reveals the child inherited two distinct homologous copies of chromosome 18 from the mother and one copy from the father. Which of the following events is the most likely cause of the child's condition?
Answer: Nondisjunction during maternal meiosis I.
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.
- A patient with Ring Chromosome 14 syndrome exhibits significant phenotypic variability, including variations in the severity of seizures and intellectual disability. This clinical variability is primarily attributed to which cytogenetic property of ring chromosomes?
Answer: The ring chromosome is mitotically unstable, leading to somatic mosaicism with cells that have lost the ring.
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.
- A prenatal amniocentesis for a fetus indicates a 47,XX,+13 karyotype in all cells analyzed. After birth, analysis of blood lymphocytes shows a mosaic karyotype: 46,XX[15]/47,XX,+13[5]. What is the most plausible explanation for this finding?
Answer: The original zygote was 47,XX,+13 and a mitotic anaphase lag led to loss of a chromosome 13.
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.
- An individual heterozygous for a large paracentric inversion on chromosome 2 undergoes meiosis. If a single crossover event occurs within the inversion loop, which of the following sets of products will be generated after Meiosis I?
Answer: One normal chromatid, one inverted chromatid, one dicentric chromatid, and one acentric fragment.
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.
- An isochromosome, such as the i(Xq) seen in some cases of Turner syndrome, is an abnormal chromosome with two identical arms. This structure is the result of which specific error during cell division?
Answer: Transverse division of the centromere during anaphase II.
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.
- A specific meiotic error occurs during spermatogenesis that results in the production of both XY sperm and sperm lacking a sex chromosome ('O' sperm) from the same primary spermatocyte. What is this event?
Answer: Nondisjunction of the X and Y chromosomes during Meiosis I.
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.
- The incidence of Down syndrome and other trisomies increases significantly with maternal age. Which molecular mechanism provides the most direct explanation for this age-related increase in Meiosis I nondisjunction?
Answer: Progressive degradation of cohesin complexes holding homologous chromosomes together.
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.
- A male is diagnosed with Klinefelter syndrome (47,XXY). Molecular analysis of the X-linked G6PD gene shows the male is heterozygous (genotype G6PD-A/G6PD-B). His mother is homozygous G6PD-A/G6PD-A, and his father is hemizygous G6PD-B. What is the parental origin of the meiotic error?
Answer: Nondisjunction during paternal Meiosis I.
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.
- A child is born with Cri-du-chat syndrome, which is caused by a terminal deletion of the short arm of chromosome 5 (5p-). Both parents have normal karyotypes. Which of the following meiotic events is a plausible de novo cause of such a terminal deletion?
Answer: Unequal crossing over between misaligned repetitive DNA sequences.
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).
- A spontaneously aborted fetus is found to have a 69,XXY karyotype. While multiple meiotic errors can cause triploidy, which of the following events is the most frequently observed underlying cause in humans?
Answer: Fertilization of a normal oocyte by two separate sperm (dispermy).
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.
- A man carries a balanced reciprocal translocation described as 46,XY,t(2;8)(q21;p12). His child presents with multiple congenital anomalies and a karyotype of 46,XY,der(8)t(2;8)(q21;p12)pat. Which statement accurately describes the genetic imbalance in the child?
Answer: Trisomy for the distal long arm of chromosome 2 and monosomy for the distal short arm of chromosome 8.
The child's karyotype indicates 46 chromosomes, including a derivative chromosome 8 inherited from the father (pat). This means the child inherited a normal chromosome 2 and the der(8) from the father, plus a normal 2 and normal 8 from the mother. The der(8) is the chromosome 8 with the distal part of chromosome 2's long arm (2q21→qter) attached. Therefore, the child has three copies of this part of chromosome 2. The child is missing the corresponding derivative chromosome 2, meaning they have only one copy of the distal part of chromosome 8's short arm (8p12→pter). This results in partial trisomy 2q and partial monosomy 8p.
- Studies of maternal Meiosis I nondisjunction events leading to trisomy 21 often find that the two homologous chromosomes that missegregated failed to undergo recombination. What is the most direct mechanistic link between this lack of crossing over and nondisjunction?
Answer: Chiasmata formed by crossovers are required to physically tether homologous chromosomes for proper spindle attachment.
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.
- A man is a carrier of a Robertsonian translocation, 45,XY,der(14;21)(q10;q10). Assuming all meiotic segregation patterns from the trivalent configuration are equally probable, what is the theoretical chance that one of his gametes will contain the necessary chromosomes to produce a zygote with translocation Down syndrome upon fertilization of a normal oocyte?
Answer: 1/6
During meiosis, the man's chromosomes 14, 21, and the der(14;21) form a trivalent. There are 6 theoretically possible gametes from segregation: (1) 14, 21 (normal); (2) der(14;21) (balanced); (3) der(14;21), 21 (causes translocation Down syndrome); (4) 14 (causes monosomy 21); (5) 21 (causes monosomy 14); (6) der(14;21), 14 (causes trisomy 14). Each is considered to have a 1/6 theoretical probability. Only the gamete containing der(14;21) and the normal chromosome 21 will result in translocation Down syndrome. Therefore, the theoretical probability is 1/6. The 1/3 risk refers to the fraction of viable offspring, which is a different calculation that excludes the lethal monosomies and trisomy 14.
