Genetics Quiz: Chromosome Structure

What this quiz covers

This quiz focuses on Chromosome Structure, 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. Metacentric
2. Submetacentric
3. Acrocentric (correct answer)
4. Holocentric

Explanation: Chromosomes are classified based on the position of the centromere. An acrocentric chromosome has its centromere located very near one end, producing a long arm (q) and a very short, often satellite-containing, arm (p). Metacentric chromosomes have a central centromere, and submetacentric chromosomes have an off-center centromere creating arms of clearly different lengths, but not as extreme as acrocentric.

Question 2

1. 46 sister chromatids and 23 centromeres
2. 92 sister chromatids and 46 centromeres (correct answer)
3. 92 sister chromatids and 92 centromeres
4. 46 sister chromatids and 46 centromeres

Explanation: A primary oocyte is a diploid (2n=46) cell that has already undergone DNA replication (S phase) before entering Meiosis I. Therefore, it contains 46 replicated chromosomes. Each replicated chromosome has one centromere, so there are 46 centromeres. Each replicated chromosome consists of two sister chromatids, so there are 46 chromosomes × 2 chromatids/chromosome = 92 sister chromatids.

Question 3

1. One fewer centromere and two fewer telomeres. (correct answer)
2. The same number of centromeres and two fewer telomeres.
3. One fewer centromere and four fewer telomeres.
4. Two fewer centromeres and two fewer telomeres.

Explanation: A standard 46,XX G1 cell has 46 chromosomes, 46 centromeres, and 46 x 2 = 92 telomeres. In a Robertsonian translocation carrier, the long arms of two acrocentric chromosomes (14 and 21) fuse, creating one large chromosome, while the small short-arm fragment is lost. This reduces the total chromosome count to 45, meaning there is one fewer centromere. The two original chromosomes had 4 telomeres in total (2 each). The new fused chromosome has only 2 telomeres. This results in a net loss of 2 telomeres. Thus, the carrier has 45 centromeres and 90 telomeres.

Question 4

1. One fewer centromere and two fewer telomeres. (correct answer)
2. The same number of centromeres and two fewer telomeres.
3. One fewer centromere and four fewer telomeres.
4. Two fewer centromeres and two fewer telomeres.

Explanation: A standard 46,XX G1 cell has 46 chromosomes, 46 centromeres, and 46 x 2 = 92 telomeres. In a Robertsonian translocation carrier, the long arms of two acrocentric chromosomes (14 and 21) fuse, creating one large chromosome, while the small short-arm fragment is lost. This reduces the total chromosome count to 45, meaning there is one fewer centromere. The two original chromosomes had 4 telomeres in total (2 each). The new fused chromosome has only 2 telomeres. This results in a net loss of 2 telomeres. Thus, the carrier has 45 centromeres and 90 telomeres.

Question 5

1. It will fail to replicate during S phase due to the missing centromere.
2. It will be degraded by cellular nucleases due to its unprotected ends.
3. It will replicate but segregate randomly, leading to its eventual loss from the cell lineage. (correct answer)
4. It will cause a permanent cell cycle arrest at the G2/M checkpoint.

Explanation: The centromere is essential for the assembly of the kinetochore, which attaches the chromosome to the mitotic spindle for proper segregation. Without a centromere, the artificial chromosome can replicate (as it has origins of replication) but cannot attach to the spindle. Consequently, its distribution to daughter cells during mitosis will be random, and it will likely be lost over subsequent cell generations.

Question 6

1. They are homologous but non-identical chromosomes.
2. They are identical chromosomes, each composed of a single chromatid. (correct answer)
3. It is a single chromosome 21 composed of two sister chromatids.
4. They are two non-homologous chromosomes of similar size.

Explanation: Meiosis II involves the separation of sister chromatids. A non-disjunction event at this stage means that a pair of sister chromatids for chromosome 21 failed to separate and moved to the same pole. These two sister chromatids, which are genetically identical (barring mutation), end up in the same spermatid. Once they are in the spermatid nucleus, they are considered two separate, identical chromosomes, each composed of a single chromatid. A Meiosis I non-disjunction would result in a gamete with two homologous (non-identical) chromosomes.

Question 7

1. Artificial chromosomes with very short synthetic alpha-satellite arrays are highly stable during mitosis. (correct answer)
2. The length of the alpha-satellite array varies significantly among different, but stable, chromosomes in the same cell.
3. A stable neocentromere is discovered that has formed on a region of the chromosome completely lacking alpha-satellite DNA.
4. Telomerase inhibition leads to centromere instability, suggesting a link between telomeres and centromeres.

Explanation: The hypothesis states that a certain length of the array is critical for function. Evidence showing that a very short, synthetic array can create a fully stable and functional centromere would directly contradict the idea that a long, natural array length is a necessary requirement. Choice B shows correlation but doesn't disprove necessity. Choice C challenges the role of the sequence itself, which is a different (though related) question. Choice D introduces a confounding factor.

Question 8

1. Telomeres would be entirely absent, as all their functions are now obsolete.
2. Telomerase activity would be extremely high to support the novel replication mechanism.
3. Telomeric DNA sequences would still be essential for chromosome capping, but telomerase would be unnecessary. (correct answer)
4. Centromere function would need to be enhanced to prevent the chromosome ends from fusing.

Explanation: Telomeres have two primary functions: 1) to be lengthened by telomerase to counteract the end-replication problem, and 2) to 'cap' the chromosome ends to prevent them from being recognized as DNA breaks and fusing. If the end-replication problem is solved, the role of telomerase becomes unnecessary. However, the capping function, which involves specific DNA sequences and associated proteins (like the shelterin complex), would still be required to maintain chromosome integrity.

Question 9

1. Telomerase would be unable to bind and extend the chromosome end.
2. The chromosome end would be recognized as a double-strand break, activating DNA repair pathways. (correct answer)
3. The centromere would become destabilized, leading to mis-segregation.
4. Replication of the lagging strand template would be blocked all along the chromosome.

Explanation: The 3' overhang is critical for forming the T-loop, a key component of the protective telomere 'cap'. Without the overhang and the subsequent T-loop, the chromosome end is exposed and resembles a DNA double-strand break (DSB). This will activate cellular DNA damage response and repair pathways, such as non-homologous end joining (NHEJ), which can lead to catastrophic end-to-end chromosome fusions. While telomerase also requires the overhang (A), the immediate threat to genomic stability is the activation of DSB repair.

Question 10

1. Prophase, where holocentric chromosomes condense into spherical, rather than linear, shapes.
2. Metaphase, where holocentric chromosomes align perpendicular to the metaphase plate.
3. Telophase, where holocentric chromosomes decondense much more rapidly than monocentric ones.
4. Anaphase, where sister chromatids separate and move to the poles as parallel bars. (correct answer)

Explanation: When you encounter questions about chromosome structure and mitotic behavior, focus on how the location of kinetochores affects chromosome movement during cell division. The key difference between monocentric and holocentric chromosomes lies in where spindle fibers attach and how this influences their appearance during mitosis. In monocentric chromosomes (like humans), kinetochores form at a single centromere, creating a point of constriction. During anaphase, sister chromatids separate and are pulled toward opposite poles by their centromeres, creating the classic V-shaped appearance as the arms trail behind. However, holocentric chromosomes have kinetochores distributed along their entire length, meaning spindle fibers attach everywhere rather than at a single point. This structural difference creates a dramatically different anaphase appearance. Since microtubules attach along the entire length of holocentric chromosomes, the separated sister chromatids move toward the poles as rigid, parallel bars rather than V-shaped structures. There's no trailing of chromosome arms because the pulling force is distributed evenly. Option A is incorrect because chromosome condensation in prophase isn't significantly affected by kinetochore distribution—both types condense into linear shapes. Option B is wrong because both chromosome types align at the metaphase plate regardless of kinetochore structure. Option C is incorrect because decondensation speed in telophase isn't determined by kinetochore distribution but by other cellular factors. Remember that kinetochore location directly determines how chromosomes move during anaphase. When you see questions about unusual chromosome structures, always consider how they would affect the mechanics of chromosome separation and movement.

Question 11

1. Arrest at the metaphase-anaphase transition, with duplicated chromosomes aligned at the cell equator. (correct answer)
2. Arrest in prophase, with chromosomes failing to condense and attach to the spindle.
3. Arrest in anaphase, with separated chromatids unable to move to the spindle poles.
4. Arrest in telophase, with a single polyploid nucleus forming.

Explanation: Separase cleaves the cohesin complexes that hold sister chromatids together. This event triggers the onset of anaphase. If separase is inhibited, cohesin remains intact, and sister chromatids cannot be pulled apart. The cell will satisfy the spindle assembly checkpoint (as chromosomes are properly attached to the spindle and aligned at the metaphase plate) but will be unable to execute the separation step. This results in an arrest at the metaphase-anaphase transition.

Question 12

1. Chromosomes align at the metaphase plate, but the spindle checkpoint arrests the cell.
2. Sister chromatids separate prematurely and are then randomly segregated to daughter cells. (correct answer)
3. Chromosomes fail to condense during prophase and cannot attach to the mitotic spindle.
4. The mitotic spindle fails to form, leading to arrest in a G2-like state with replicated DNA.

Explanation: Proper cohesion is essential for holding sister chromatids together until anaphase, which ensures their bipolar attachment to the mitotic spindle. If cohesion is weak, sister chromatids may separate before metaphase. This premature separation prevents stable bipolar attachment and proper alignment. The individual chromatids will then be segregated randomly and unequally into the daughter cells, leading to severe aneuploidy.

Question 13

1. 40
2. 80 (correct answer)
3. 160
4. 20

Explanation: The somatic cell is diploid with 2n=40, so n=20. After Meiosis I, the cell is haploid (n=20), but each of the 20 chromosomes still consists of two sister chromatids. A replicated chromosome has four telomeres (one at the end of each of the four 'arms' of the two chromatids). Therefore, the total number of telomeres is 20 chromosomes × 4 telomeres/chromosome = 80.

Question 14

1. Arrest at the metaphase-anaphase transition, with duplicated chromosomes aligned at the cell equator. (correct answer)
2. Arrest in prophase, with chromosomes failing to condense and attach to the spindle.
3. Arrest in anaphase, with separated chromatids unable to move to the spindle poles.
4. Arrest in telophase, with a single polyploid nucleus forming.

Explanation: Separase cleaves the cohesin complexes that hold sister chromatids together. This event triggers the onset of anaphase. If separase is inhibited, cohesin remains intact, and sister chromatids cannot be pulled apart. The cell will satisfy the spindle assembly checkpoint (as chromosomes are properly attached to the spindle and aligned at the metaphase plate) but will be unable to execute the separation step. This results in an arrest at the metaphase-anaphase transition.

Question 15

1. Immediate end-to-end fusion of chromosomes in the first mitotic division.
2. Failure of DNA replication initiation at the origins along the chromosome.
3. Progressive shortening of chromosome ends with each cell division, leading to senescence. (correct answer)
4. Destabilization of centromeres, resulting in widespread aneuploidy.

Explanation: The RNA component of telomerase serves as the template for adding repetitive DNA sequences to the ends of chromosomes. Without this template, the telomerase enzyme is non-functional. In dividing somatic cells, this leads to the 'end-replication problem,' where chromosomes become progressively shorter with each replication cycle. This eventual shortening will lead to the loss of genetic information and trigger cellular senescence or apoptosis.

Question 16

1. 40
2. 80 (correct answer)
3. 160
4. 20

Explanation: The somatic cell is diploid with 2n=40, so n=20. After Meiosis I, the cell is haploid (n=20), but each of the 20 chromosomes still consists of two sister chromatids. A replicated chromosome has four telomeres (one at the end of each of the four 'arms' of the two chromatids). Therefore, the total number of telomeres is 20 chromosomes × 4 telomeres/chromosome = 80.

Question 17

1. Metacentric
2. Submetacentric
3. Acrocentric (correct answer)
4. Holocentric

Explanation: Chromosomes are classified based on the position of the centromere. An acrocentric chromosome has its centromere located very near one end, producing a long arm (q) and a very short, often satellite-containing, arm (p). Metacentric chromosomes have a central centromere, and submetacentric chromosomes have an off-center centromere creating arms of clearly different lengths, but not as extreme as acrocentric.

Question 18

1. Telomeres would be entirely absent, as all their functions are now obsolete.
2. Telomerase activity would be extremely high to support the novel replication mechanism.
3. Telomeric DNA sequences would still be essential for chromosome capping, but telomerase would be unnecessary. (correct answer)
4. Centromere function would need to be enhanced to prevent the chromosome ends from fusing.

Explanation: Telomeres have two primary functions: 1) to be lengthened by telomerase to counteract the end-replication problem, and 2) to 'cap' the chromosome ends to prevent them from being recognized as DNA breaks and fusing. If the end-replication problem is solved, the role of telomerase becomes unnecessary. However, the capping function, which involves specific DNA sequences and associated proteins (like the shelterin complex), would still be required to maintain chromosome integrity.

Question 19

1. Telomerase would be unable to bind and extend the chromosome end.
2. The chromosome end would be recognized as a double-strand break, activating DNA repair pathways. (correct answer)
3. The centromere would become destabilized, leading to mis-segregation.
4. Replication of the lagging strand template would be blocked all along the chromosome.

Explanation: The 3' overhang is critical for forming the T-loop, a key component of the protective telomere 'cap'. Without the overhang and the subsequent T-loop, the chromosome end is exposed and resembles a DNA double-strand break (DSB). This will activate cellular DNA damage response and repair pathways, such as non-homologous end joining (NHEJ), which can lead to catastrophic end-to-end chromosome fusions. While telomerase also requires the overhang (A), the immediate threat to genomic stability is the activation of DSB repair.

Question 20

1. DNA = 2C; Centromeres = 2N; Telomeres = 8N (correct answer)
2. DNA = 2C; Centromeres = 4N; Telomeres = 8N
3. DNA = C; Centromeres = 2N; Telomeres = 4N
4. DNA = 2C; Centromeres = 2N; Telomeres = 4N

Explanation: In G1, the cell has DNA content C, 2N chromosomes, 2N centromeres, and 2N x 2 = 4N telomeres. After S phase, the DNA content doubles to 2C. In metaphase, the cell still has 2N chromosomes (each now consisting of two chromatids), so there are still 2N centromeres. However, each of the 2N replicated chromosomes has 4 telomeres (one at each chromatid end). Therefore, the total number of telomeres is 2N chromosomes x 4 telomeres/chromosome = 8N.