Genetics Quiz: Karyotypes And Aneuploidies

What this quiz covers

This quiz focuses on Karyotypes And Aneuploidies, 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. The X chromosome contains fewer protein-coding genes than chromosome 21.
2. Dosage compensation via X-chromosome inactivation mitigates the effect of the extra X chromosome. (correct answer)
3. Autosomes are more susceptible to nondisjunction than sex chromosomes.
4. The extra X chromosome is typically eliminated from most somatic cells during early development.

Explanation: In mammals, dosage compensation is achieved by inactivating all but one X chromosome in each somatic cell. In a 47,XXX individual, two of the three X chromosomes are inactivated and form Barr bodies. This silences most of the genes on the extra X chromosomes, largely normalizing the gene dosage and leading to a milder phenotype. There is no equivalent mechanism for silencing an entire extra autosome, so the gene dosage imbalance in Trisomy 21 is not corrected, resulting in a more severe clinical syndrome.

Question 2

1. Only ova containing a single X chromosome (23,X).
2. Ova containing either one X (23,X) or two X chromosomes (24,XX). (correct answer)
3. Ova containing either two X (24,XX) or no X chromosomes (22,O).
4. Ova containing one X (23,X), two X (24,XX), or three X chromosomes (25,XXX).

Explanation: The segregation pattern described (two X chromosomes to one pole, one to the other) means that the secondary oocytes will contain either two X chromosomes or one X chromosome. After a normal meiosis II, these will mature into ova containing either two X chromosomes (n+1, or 24,XX) or one X chromosome (n, or 23,X). Therefore, she can produce both normal and aneuploid gametes.

Question 3

1. Nondisjunction during meiosis I.
2. Nondisjunction during meiosis II. (correct answer)
3. Nondisjunction during a pre-meiotic mitotic division.
4. Anaphase lag of a chromosome during meiosis I.

Explanation: Nondisjunction in meiosis I affects the separation of homologous chromosomes, resulting in all four final gametes being aneuploid (two n+1, two n-1). In contrast, nondisjunction in meiosis II affects the separation of sister chromatids. This error occurs in only one of the two cells that result from meiosis I. The other cell proceeds through meiosis II normally. Consequently, the final products are two normal (n) gametes and two aneuploid gametes (one n+1, one n-1).

Question 4

1. All pollen will have 12 chromosomes.
2. All pollen will have 13 chromosomes.
3. Pollen will have either 12 or 13 chromosomes. (correct answer)
4. Pollen will have either 12, 12.5, or 13 chromosomes.

Explanation: The plant is 2n+1, so it has 25 chromosomes. During meiosis, the 11 normal bivalents will segregate to produce gametes with 11 chromosomes each. The trivalent of chromosome 5 will segregate such that one secondary meiocyte receives one copy of chromosome 5 and the other receives two copies. After meiosis II, the resulting pollen grains will therefore contain either 11+1=12 chromosomes (normal haploid number, n) or 11+2=13 chromosomes (n+1).

Question 5

1. Nondisjunction during meiosis I.
2. Nondisjunction during meiosis II. (correct answer)
3. Nondisjunction during a pre-meiotic mitotic division.
4. Anaphase lag of a chromosome during meiosis I.

Explanation: Nondisjunction in meiosis I affects the separation of homologous chromosomes, resulting in all four final gametes being aneuploid (two n+1, two n-1). In contrast, nondisjunction in meiosis II affects the separation of sister chromatids. This error occurs in only one of the two cells that result from meiosis I. The other cell proceeds through meiosis II normally. Consequently, the final products are two normal (n) gametes and two aneuploid gametes (one n+1, one n-1).

Question 6

1. The X chromosome contains fewer protein-coding genes than chromosome 21.
2. Dosage compensation via X-chromosome inactivation mitigates the effect of the extra X chromosome. (correct answer)
3. Autosomes are more susceptible to nondisjunction than sex chromosomes.
4. The extra X chromosome is typically eliminated from most somatic cells during early development.

Explanation: In mammals, dosage compensation is achieved by inactivating all but one X chromosome in each somatic cell. In a 47,XXX individual, two of the three X chromosomes are inactivated and form Barr bodies. This silences most of the genes on the extra X chromosomes, largely normalizing the gene dosage and leading to a milder phenotype. There is no equivalent mechanism for silencing an entire extra autosome, so the gene dosage imbalance in Trisomy 21 is not corrected, resulting in a more severe clinical syndrome.

Question 7

1. It is a structural aberration, whereas aneuploidy is a numerical aberration. (correct answer)
2. It involves a gain of genetic material, whereas aneuploidy involves a loss.
3. It is always acquired post-zygotically, whereas aneuploidy is always meiotic in origin.
4. It affects an autosome, whereas aneuploidy only affects sex chromosomes.

Explanation: The notation r(7) signifies a ring chromosome 7, which is formed when the chromosome breaks at both ends (p22 and q36) and the broken ends fuse. This is a change in the chromosome's structure. The total chromosome count is 46, which is normal. Aneuploidy, such as Turner syndrome (45,X), is defined by an abnormal number of chromosomes. This is the fundamental difference between the two types of abnormalities.

Question 8

1. Two spermatids with 24 chromosomes (n+1), and two spermatids with 22 chromosomes (n-1). (correct answer)
2. Two normal spermatids with 23 chromosomes (n), one with 24 (n+1), and one with 22 (n-1).
3. Four spermatids, all with 24 chromosomes (n+1).
4. One normal spermatid (n), two with 24 chromosomes (n+1), and one nullisomic (n-1).

Explanation: Nondisjunction in meiosis I means the homologous pair of chromosome 21 fails to separate. One secondary spermatocyte receives both homologs, and the other receives none. The secondary spermatocyte with both homologs proceeds through meiosis II to produce two spermatids, each with an extra chromosome 21 (n+1, or 24 total). The other secondary spermatocyte with no chromosome 21 produces two spermatids that are missing chromosome 21 (n-1, or 22 total). Therefore, all four products are aneuploid.

Question 9

1. Unmasking of recessive lethal alleles on the single autosome is the primary cause of embryonic lethality.
2. Autosomal monosomies prevent proper implantation in the uterus, whereas trisomies do not.
3. Haploinsufficiency of numerous genes on the missing chromosome creates an intolerable developmental burden. (correct answer)
4. The cell cycle checkpoints are more sensitive to the loss of a chromosome than to the gain of one.

Explanation: Haploinsufficiency refers to a situation where a single copy of a gene is insufficient to provide the normal function. The loss of an entire autosome results in haploinsufficiency for hundreds or thousands of genes simultaneously. This massive disruption of gene dosage is generally incompatible with embryonic development. While unmasking of recessive lethal alleles (A) can contribute, haploinsufficiency is considered the more fundamental and universal reason for the lethality of autosomal monosomies.

Question 10

1. Paternal meiosis I nondisjunction leading to a trisomy 16 zygote, followed by post-zygotic loss of the maternal chromosome 16. (correct answer)
2. Fertilization of a normal ovum by a disomic sperm (containing two chromosome 16s).
3. Maternal meiosis I nondisjunction leading to a trisomy 16 zygote, followed by post-zygotic loss of a maternal chromosome 16.
4. Post-zygotic nondisjunction in a normal 46,XX zygote, leading to trisomic and monosomic cell lines.

Explanation: The final result is paternal uniparental disomy (UPD) for chromosome 16. The finding of trisomy 16 in the placenta suggests the zygote was originally trisomic. This conceptus then underwent 'trisomy rescue' where one of the three chromosomes 16 was lost in the cell lineage that forms the fetus. For the child to end up with two paternal chromosomes 16, the original trisomic zygote must have had two paternal copies and one maternal copy. This would result from paternal meiotic nondisjunction. The subsequent loss of the single maternal chromosome 16 would result in a 46,XX karyotype with paternal UPD 16.

Question 11

1. Aneuploidy, 69,XXY
2. Trisomy, 47,XXY,+1...+22
3. Polyploidy, 69,XXY (correct answer)
4. Mosaicism, 46,XY/69,XXY

Explanation: The cell has 69 chromosomes, which is exactly three times the haploid number (3n=69). This condition is known as polyploidy, specifically triploidy. The sex chromosomes are XXY, which is consistent with three sets of chromosomes. Aneuploidy refers to the gain or loss of one or more individual chromosomes, not entire sets. Trisomy refers to having three copies of a single specific chromosome (e.g., 47,XX,+21). Mosaicism would imply the presence of at least two different cell lines, which is not stated.

Question 12

1. The father's translocation is the cause of the first child's Down syndrome.
2. The first child's Down syndrome resulted from a maternal meiotic error, so there is no increased risk of recurrence.
3. The first child's condition is due to a sporadic meiotic error, but the couple has a high risk of recurrence due to the paternal translocation. (correct answer)
4. The first child's karyotype is incompatible with the father's karyotype, suggesting non-paternity.

Explanation: The first child has standard Trisomy 21 (47,XX,+21), characterized by three separate copies of chromosome 21. This is distinct from translocation Down syndrome, which would have a karyotype of 46,XX,der(14;21),+21. Therefore, the first child's condition was caused by a de novo nondisjunction event, unrelated to the father's translocation. However, the father is a balanced carrier of a Robertsonian translocation involving chromosome 21. This places the couple at a significantly increased risk (empirically 10-15%) of having a future child with translocation Down syndrome.

Question 13

1. 47,XY,+13, due to nondisjunction in maternal meiosis I.
2. 47,XY,+13, due to nondisjunction in maternal meiosis II. (correct answer)
3. 46,XY,der(13;14)(q10;q10),+13, inherited from a carrier parent.
4. 47,XY,+13, due to nondisjunction in paternal meiosis II.

Explanation: The infant has Patau syndrome, so the karyotype is 47,XY,+13. The marker analysis shows two identical maternal chromosomes 13. Nondisjunction in meiosis I involves the failure of homologous chromosomes to separate, which would result in the infant receiving two different (heterozygous) chromosomes from the mother. Nondisjunction in meiosis II involves the failure of sister chromatids to separate. Since sister chromatids are identical (barring crossing over away from the centromere), receiving both sister chromatids results in homozygosity for centromeric markers. Therefore, the error occurred during maternal meiosis II.

Question 14

1. 1/2
2. 1/3 (correct answer)
3. 1/4
4. Approximately 10-15%

Explanation: A carrier of a Robertsonian translocation between chromosomes 14 and 21 produces several types of gametes. There are three potential combinations that can lead to a viable offspring after fertilization by a normal gamete: (1) a gamete with a normal chromosome 14 and 21 (leading to a normal child), (2) a gamete with the der(14;21) chromosome (leading to a balanced carrier like the mother), and (3) a gamete with the der(14;21) chromosome plus a normal chromosome 21 (leading to translocation Down syndrome). Other gamete types lead to non-viable monosomies or trisomies. Therefore, of the three viable outcomes, one results in Down syndrome, giving a theoretical risk of 1/3. The value of 10-15% represents the lower, empirically observed risk.

Question 15

1. 47,XY,+13, due to nondisjunction in maternal meiosis I.
2. 47,XY,+13, due to nondisjunction in maternal meiosis II. (correct answer)
3. 46,XY,der(13;14)(q10;q10),+13, inherited from a carrier parent.
4. 47,XY,+13, due to nondisjunction in paternal meiosis II.

Explanation: The infant has Patau syndrome, so the karyotype is 47,XY,+13. The marker analysis shows two identical maternal chromosomes 13. Nondisjunction in meiosis I involves the failure of homologous chromosomes to separate, which would result in the infant receiving two different (heterozygous) chromosomes from the mother. Nondisjunction in meiosis II involves the failure of sister chromatids to separate. Since sister chromatids are identical (barring crossing over away from the centromere), receiving both sister chromatids results in homozygosity for centromeric markers. Therefore, the error occurred during maternal meiosis II.

Question 16

1. Only ova containing a single X chromosome (23,X).
2. Ova containing either one X (23,X) or two X chromosomes (24,XX). (correct answer)
3. Ova containing either two X (24,XX) or no X chromosomes (22,O).
4. Ova containing one X (23,X), two X (24,XX), or three X chromosomes (25,XXX).

Explanation: The segregation pattern described (two X chromosomes to one pole, one to the other) means that the secondary oocytes will contain either two X chromosomes or one X chromosome. After a normal meiosis II, these will mature into ova containing either two X chromosomes (n+1, or 24,XX) or one X chromosome (n, or 23,X). Therefore, she can produce both normal and aneuploid gametes.

Question 17

1. All pollen will have 12 chromosomes.
2. All pollen will have 13 chromosomes.
3. Pollen will have either 12 or 13 chromosomes. (correct answer)
4. Pollen will have either 12, 12.5, or 13 chromosomes.

Explanation: The plant is 2n+1, so it has 25 chromosomes. During meiosis, the 11 normal bivalents will segregate to produce gametes with 11 chromosomes each. The trivalent of chromosome 5 will segregate such that one secondary meiocyte receives one copy of chromosome 5 and the other receives two copies. After meiosis II, the resulting pollen grains will therefore contain either 11+1=12 chromosomes (normal haploid number, n) or 11+2=13 chromosomes (n+1).

Question 18

1. Unmasking of recessive lethal alleles on the single autosome is the primary cause of embryonic lethality.
2. Autosomal monosomies prevent proper implantation in the uterus, whereas trisomies do not.
3. Haploinsufficiency of numerous genes on the missing chromosome creates an intolerable developmental burden. (correct answer)
4. The cell cycle checkpoints are more sensitive to the loss of a chromosome than to the gain of one.

Explanation: Haploinsufficiency refers to a situation where a single copy of a gene is insufficient to provide the normal function. The loss of an entire autosome results in haploinsufficiency for hundreds or thousands of genes simultaneously. This massive disruption of gene dosage is generally incompatible with embryonic development. While unmasking of recessive lethal alleles (A) can contribute, haploinsufficiency is considered the more fundamental and universal reason for the lethality of autosomal monosomies.

Question 19

1. Aneuploidy, 69,XXY
2. Trisomy, 47,XXY,+1...+22
3. Polyploidy, 69,XXY (correct answer)
4. Mosaicism, 46,XY/69,XXY

Explanation: The cell has 69 chromosomes, which is exactly three times the haploid number (3n=69). This condition is known as polyploidy, specifically triploidy. The sex chromosomes are XXY, which is consistent with three sets of chromosomes. Aneuploidy refers to the gain or loss of one or more individual chromosomes, not entire sets. Trisomy refers to having three copies of a single specific chromosome (e.g., 47,XX,+21). Mosaicism would imply the presence of at least two different cell lines, which is not stated.

Question 20

1. It is a structural aberration, whereas aneuploidy is a numerical aberration. (correct answer)
2. It involves a gain of genetic material, whereas aneuploidy involves a loss.
3. It is always acquired post-zygotically, whereas aneuploidy is always meiotic in origin.
4. It affects an autosome, whereas aneuploidy only affects sex chromosomes.

Explanation: The notation r(7) signifies a ring chromosome 7, which is formed when the chromosome breaks at both ends (p22 and q36) and the broken ends fuse. This is a change in the chromosome's structure. The total chromosome count is 46, which is normal. Aneuploidy, such as Turner syndrome (45,X), is defined by an abnormal number of chromosomes. This is the fundamental difference between the two types of abnormalities.