Genetics Quiz: Chromosomal Rearrangements
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Chromosomal RearrangementsQuestion 1 of 20

The Bar eye phenotype in Drosophila is a classic example of a trait influenced by gene dosage. The phenotype's severity is directly proportional to the number of copies of the 16A region on the X chromosome. Unequal crossing over in a female homozygous for the Bar mutation can produce gametes that lead to offspring with either a wild-type eye or an extreme 'double-Bar' phenotype. What type of rearrangement is the Bar mutation itself?

A pericentric inversion that disrupts a gene in region 16A.
A tandem duplication of region 16A.
An interstitial deletion of a repressor element near region 16A.
A position effect caused by translocating region 16A near heterochromatin.
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Genetics Quiz

Genetics Quiz: Chromosomal Rearrangements

Practice Chromosomal Rearrangements in Genetics with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

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

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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.

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Question 1

The Bar eye phenotype in Drosophila is a classic example of a trait influenced by gene dosage. The phenotype's severity is directly proportional to the number of copies of the 16A region on the X chromosome. Unequal crossing over in a female homozygous for the Bar mutation can produce gametes that lead to offspring with either a wild-type eye or an extreme 'double-Bar' phenotype. What type of rearrangement is the Bar mutation itself?

  1. A pericentric inversion that disrupts a gene in region 16A.
  2. A tandem duplication of region 16A. (correct answer)
  3. An interstitial deletion of a repressor element near region 16A.
  4. A position effect caused by translocating region 16A near heterochromatin.

Explanation: The graded effect of the Bar phenotype, where more copies of region 16A lead to a more severe phenotype, is characteristic of a gene dosage effect caused by duplication. The original Bar mutation is a tandem duplication of the 16A region. Unequal crossing over between the duplicated segments in a Bar homozygote can then generate a chromatid with a single copy (reverting to wild-type) and a chromatid with three copies (a triplication, causing the 'double-Bar' phenotype). Deletions or inversions do not explain the copy number-dependent severity and the generation of both wild-type and double-Bar.

Question 2

In Drosophila, an inversion moves the wild-type white gene (w⁺) from a euchromatic region to a new location adjacent to centromeric heterochromatin. This results in flies with mottled eyes, containing both red and white patches. This variable gene expression due to a change in chromosomal position is a classic example of:

  1. Pseudodominance
  2. Haploinsufficiency
  3. Dosage compensation
  4. Position effect variegation (correct answer)

Explanation: Position effect variegation (PEV) occurs when a gene's expression is altered by its new chromosomal location, typically when moved from euchromatin (transcriptionally active) to heterochromatin (transcriptionally repressed). The heterochromatic state can spread into the relocated gene, silencing it in some cells but not others, leading to a variegated or mosaic phenotype. This is distinct from pseudodominance (unmasking a recessive allele), haploinsufficiency (a dosage effect from deletion), and dosage compensation (equalizing gene expression between sexes).

Question 3

A chromosome has gene loci in the order cen-P-Q-R-S-T. Cytogenetic analysis of a phenotypically normal individual reveals a rearranged homologous chromosome with the order cen-P-S-R-Q-T. What is the most likely consequence for this individual's reproductive fitness?

  1. No significant consequence, as the rearrangement is balanced and all genes are present.
  2. Reduced fertility due to the production of inviable gametes resulting from crossover within the rearranged segment. (correct answer)
  3. High probability of producing offspring with phenotypes related to the haploinsufficiency of genes Q, R, and S.
  4. Reduced fertility due to the production of viable but unbalanced gametes containing duplications and deletions.

Explanation: The rearrangement is an inversion of the Q-R-S segment. Since it does not include the centromere (cen), it is a paracentric inversion. While the carrier is phenotypically normal because the inversion is balanced (A is incorrect), they may experience reduced fertility. A crossover within the Q-R-S inversion loop during meiosis produces dicentric and acentric chromatids, which lead to genetically unbalanced and inviable gametes (B is correct). This effectively suppresses recombination in the region. Unlike pericentric inversions, paracentric inversions do not typically produce viable but unbalanced gametes (D is incorrect). There is no loss of genes, so haploinsufficiency (C) is not the direct cause.

Question 4

Williams-Beuren syndrome is a congenital disorder caused by the loss of approximately 27 genes, including the elastin gene (ELN), from one copy of chromosome 7. The phenotype is a direct result of having only a single functional copy of these genes. This genetic situation is best described as:

  1. Codominance
  2. Haploinsufficiency (correct answer)
  3. Pseudodominance
  4. Position effect variegation

Explanation: Haploinsufficiency occurs when a diploid organism has only a single functional copy of a gene (with the other copy inactivated or lost) and the single functional copy does not produce enough of a gene product to bring about a wild-type condition, leading to an abnormal phenotype. This perfectly describes the mechanism of Williams-Beuren syndrome. Codominance refers to the expression of both alleles. Pseudodominance is the expression of a recessive allele due to deletion of the dominant allele. Position effect variegation is variable expression due to a change in the gene's chromosomal environment.

Question 5

The Bar eye phenotype in Drosophila is a classic example of a trait influenced by gene dosage. The phenotype's severity is directly proportional to the number of copies of the 16A region on the X chromosome. Unequal crossing over in a female homozygous for the Bar mutation can produce gametes that lead to offspring with either a wild-type eye or an extreme 'double-Bar' phenotype. What type of rearrangement is the Bar mutation itself?

  1. A pericentric inversion that disrupts a gene in region 16A.
  2. A tandem duplication of region 16A. (correct answer)
  3. An interstitial deletion of a repressor element near region 16A.
  4. A position effect caused by translocating region 16A near heterochromatin.

Explanation: The graded effect of the Bar phenotype, where more copies of region 16A lead to a more severe phenotype, is characteristic of a gene dosage effect caused by duplication. The original Bar mutation is a tandem duplication of the 16A region. Unequal crossing over between the duplicated segments in a Bar homozygote can then generate a chromatid with a single copy (reverting to wild-type) and a chromatid with three copies (a triplication, causing the 'double-Bar' phenotype). Deletions or inversions do not explain the copy number-dependent severity and the generation of both wild-type and double-Bar.

Question 6

A researcher studies a true-breeding Drosophila stock with dominant wild-type alleles for five linked genes (A, B, C, D, E). This stock is crossed to a stock homozygous for the recessive alleles (a, b, c, d, e). The resulting F1 flies, which are heterozygous at all five loci, are then test-crossed. The analysis of thousands of F2 progeny reveals that a small, distinct class of flies unexpectedly expresses the recessive phenotypes for both traits C and D, despite inheriting the chromosome from the wild-type parent. These flies also exhibit reduced viability.

Based on the passage, what is the most probable chromosomal arrangement in the F1 parent that explains the expression of the recessive c and d alleles?

  1. A spontaneous reversion of the C and D alleles to c and d.
  2. A balanced reciprocal translocation with a breakpoint between B and C.
  3. An interstitial deletion on the wild-type chromosome spanning the C and D loci. (correct answer)
  4. Epigenetic silencing of the C and D alleles on the wild-type chromosome.

Explanation: The expression of recessive alleles (c and d) when a dominant allele is expected to be present is known as pseudodominance. This phenomenon occurs when the segment of the homologous chromosome containing the dominant alleles is lost due to a deletion. The F1 fly has one chromosome with A-B-C-D-E and another with a-b-c-d-e. If the C and D loci are deleted from the first chromosome, the recessive c and d alleles on the second chromosome will be expressed. This deletion would also explain the reduced viability observed. A spontaneous reversion of two specific alleles simultaneously (A) is extremely improbable. A balanced translocation (B) would not remove the genes. Epigenetic silencing (D) is a possibility but a deletion is a more direct and classic cytogenetic explanation for pseudodominance of linked genes.

Question 7

Certain genetic disorders, such as Charcot-Marie-Tooth disease type 1A, are caused by the duplication of a specific gene region (e.g., PMP22 on chromosome 17). This region is known to be flanked by low-copy repeats (LCRs). What is the most common meiotic mechanism responsible for generating such a duplication?

  1. Unequal crossing over between misaligned homologous LCRs. (correct answer)
  2. Non-homologous end joining following a chromosome break.
  3. Replication slippage during S phase in a germline stem cell.
  4. Nondisjunction of sister chromatids during meiosis II.

Explanation: When you encounter questions about gene duplications flanked by low-copy repeats (LCRs), think about how repetitive DNA sequences can cause problems during meiosis. LCRs are nearly identical DNA sequences that can confuse the cellular machinery responsible for chromosome pairing and crossing over. The correct mechanism is unequal crossing over between misaligned homologous LCRs (answer A). During meiosis, homologous chromosomes normally align perfectly before crossing over. However, when LCRs are present, the similar sequences can cause misalignment—one chromosome's LCR pairs with the wrong LCR on its homolog. When crossing over occurs between these misaligned repeats, one chromosome gains the duplicated region (causing Charcot-Marie-Tooth disease type 1A) while the other loses it (causing a deletion syndrome). Answer B, non-homologous end joining, repairs double-strand breaks but doesn't typically generate duplications of specific gene regions flanked by repeats. Answer C, replication slippage, can cause small insertions or deletions but isn't the primary mechanism for large duplications involving LCRs spanning entire genes. Answer D, nondisjunction of sister chromatids, would cause aneuploidy (abnormal chromosome numbers) rather than structural rearrangements like duplications. Study tip: Remember that LCRs are "troublemakers" during meiosis because they create opportunities for misalignment. When you see questions about genomic disorders involving duplications or deletions with LCRs mentioned, think unequal crossing over first. This mechanism explains many common genomic disorders beyond Charcot-Marie-Tooth disease.

Question 8

During prophase I of meiosis, the homologous chromosomes of an individual heterozygous for a chromosomal rearrangement form a characteristic cross-shaped quadrivalent structure. This individual is phenotypically normal but has a history of semi-sterility. This meiotic configuration is indicative of which type of rearrangement?

  1. A large pericentric inversion
  2. A balanced reciprocal translocation (correct answer)
  3. A Robertsonian translocation
  4. A large interstitial duplication

Explanation: A cross-shaped quadrivalent is the characteristic pairing configuration formed during meiosis in an individual heterozygous for a balanced reciprocal translocation. This structure allows for the alignment of all homologous regions across the two pairs of chromosomes involved. Inversions (A) and duplications (D) form loops, while a Robertsonian translocation (C) typically forms a trivalent. The semi-sterility is explained by the production of unbalanced gametes through adjacent segregation from this quadrivalent.

Question 9

A researcher uses fluorescence in situ hybridization (FISH) with a red probe for the 5q terminus and a green probe for the 13q terminus. In a metaphase spread from a phenotypically normal patient, most cells show two red signals and two green signals on four separate chromosomes. However, the researcher notes one chromosome 5 with a red signal, one chromosome 13 with a green signal, and an abnormal chromosome showing both a red and a green signal. What is the most likely rearrangement?

  1. A pericentric inversion on chromosome 5.
  2. An interstitial deletion on chromosome 13.
  3. A balanced reciprocal translocation between chromosomes 5 and 13. (correct answer)
  4. Trisomy for a portion of chromosome 5 and monosomy for a portion of chromosome 13.

Explanation: The observation of a single chromosome with signals from two different chromosomes (red from 5qter and green from 13qter) indicates that parts of chromosome 5 and 13 have been joined. Since the patient is phenotypically normal, the rearrangement is likely balanced. A balanced reciprocal translocation, where the terminal segments of 5q and 13q were exchanged, would produce exactly this FISH pattern: a normal 5, a normal 13, a derivative 5 carrying the end of 13, and a derivative 13 carrying the end of 5. The abnormal chromosome seen is one of these derivative chromosomes. The other choices would not produce a chromosome with both red and green signals.

Question 10

The Philadelphia chromosome is a cytogenetic abnormality strongly associated with chronic myelogenous leukemia (CML). It results in the formation of a novel fusion gene, BCR-ABL, whose protein product is a constitutively active tyrosine kinase. The BCR gene is on chromosome 22, and the ABL gene is on chromosome 9. What specific type of rearrangement creates this oncogenic fusion?

  1. A nonreciprocal translocation of ABL into the BCR locus.
  2. A pericentric inversion on chromosome 22 with a breakpoint in BCR.
  3. A reciprocal translocation between the long arms of chromosomes 9 and 22. (correct answer)
  4. A Robertsonian translocation involving chromosomes 9 and 22.

Explanation: The Philadelphia chromosome is the result of a specific balanced reciprocal translocation, denoted t(9;22)(q34;q11). The distal portion of the long arm of chromosome 9, containing the ABL gene, is translocated to chromosome 22, where it fuses with the BCR gene. Concurrently, a piece of chromosome 22 moves to chromosome 9. A nonreciprocal translocation (A) is an incomplete description. An inversion (B) would not move a gene from chromosome 9. A Robertsonian translocation (D) only occurs between acrocentric chromosomes, which 9 and 22 are not.

Question 11

A phenotypically normal woman has a child with Down syndrome. Karyotyping reveals the child has 46 chromosomes, including one normal chromosome 14, one normal chromosome 21, and a derivative chromosome consisting of the long arms of chromosomes 14 and 21. Given that this condition can be familial, what is the most likely karyotype of the phenotypically normal mother?

  1. 47,XX,+21
  2. 46,XX,t(14;21)(q11;q11)
  3. 45,XX,der(14;21)(q10;q10) (correct answer)
  4. 46,XX/47,XX,+21 mosaic

Explanation: The child has translocation Down syndrome with 46 chromosomes. This occurs when a child inherits a normal chromosome 21 from one parent and both a normal chromosome 14 and a der(14;21) translocation chromosome from a carrier parent. The carrier parent is phenotypically normal because the translocation is balanced. They have a total of 45 chromosomes, including the fused der(14;21) chromosome, but have the correct amount of genetic material. A karyotype of 45,XX,der(14;21) correctly describes such a female carrier. Choice A is the karyotype for trisomy 21, which the mother does not have. Choice B describes a reciprocal translocation, but a Robertsonian translocation carrier has only 45 chromosomes. Choice D describes mosaicism for trisomy 21, which is not the typical cause of familial Down syndrome.

Question 12

For a carrier of a balanced reciprocal translocation, meiotic segregation of the four chromosomes in the quadrivalent determines the genetic content of the gametes. Which segregation pattern is responsible for producing genetically unbalanced gametes containing one normal and one translocated chromosome?

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

Explanation: In adjacent-1 segregation, homologous centromeres separate at anaphase I, but each pole receives one normal and one translocated chromosome. This results in gametes that are unbalanced, with a duplication of some genetic material and a deletion of other material. In contrast, alternate segregation (A) sends both normal chromosomes to one pole and both translocated chromosomes to the other, producing balanced gametes (either fully normal or balanced carrier). Somatic segregation (C) is not a standard meiotic term, and segregation is not random (D).

Question 13

A researcher studies a true-breeding Drosophila stock with dominant wild-type alleles for five linked genes (A, B, C, D, E). This stock is crossed to a stock homozygous for the recessive alleles (a, b, c, d, e). The resulting F1 flies, which are heterozygous at all five loci, are then test-crossed. The analysis of thousands of F2 progeny reveals that a small, distinct class of flies unexpectedly expresses the recessive phenotypes for both traits C and D, despite inheriting the chromosome from the wild-type parent. These flies also exhibit reduced viability.

Based on the passage, what is the most probable chromosomal arrangement in the F1 parent that explains the expression of the recessive c and d alleles?

  1. A spontaneous reversion of the C and D alleles to c and d.
  2. A balanced reciprocal translocation with a breakpoint between B and C.
  3. An interstitial deletion on the wild-type chromosome spanning the C and D loci. (correct answer)
  4. Epigenetic silencing of the C and D alleles on the wild-type chromosome.

Explanation: The expression of recessive alleles (c and d) when a dominant allele is expected to be present is known as pseudodominance. This phenomenon occurs when the segment of the homologous chromosome containing the dominant alleles is lost due to a deletion. The F1 fly has one chromosome with A-B-C-D-E and another with a-b-c-d-e. If the C and D loci are deleted from the first chromosome, the recessive c and d alleles on the second chromosome will be expressed. This deletion would also explain the reduced viability observed. A spontaneous reversion of two specific alleles simultaneously (A) is extremely improbable. A balanced translocation (B) would not remove the genes. Epigenetic silencing (D) is a possibility but a deletion is a more direct and classic cytogenetic explanation for pseudodominance of linked genes.

Question 14

An individual is heterozygous for a large paracentric inversion. If a single crossover event occurs within the inversion loop during meiosis I, what is the fate of the chromosomal products at the completion of meiosis?

  1. Two viable gametes with parental gene order and two viable gametes with balanced, recombinant gene order.
  2. Four viable gametes, all of which contain chromosomes with duplications and deletions.
  3. Two viable gametes with parental gene order, one non-viable gamete with a dicentric chromosome, and one non-viable gamete with an acentric fragment. (correct answer)
  4. Two viable gametes with parental gene order and two non-viable aneuploid gametes resulting from nondisjunction.

Explanation: A crossover within a paracentric inversion loop produces four chromatids: two non-recombinant parental chromatids (which are normal and result in viable gametes), one dicentric chromatid (which forms a bridge at anaphase I and breaks), and one acentric fragment (which is lost because it cannot attach to the spindle). The gametes receiving the broken dicentric chromosome or the one lacking the acentric fragment are genetically unbalanced and non-viable. Therefore, the only viable products are the parental ones. Choice C provides the most accurate and complete description of this outcome.

Question 15

During prophase I of meiosis, the homologous chromosomes of an individual heterozygous for a chromosomal rearrangement form a characteristic cross-shaped quadrivalent structure. This individual is phenotypically normal but has a history of semi-sterility. This meiotic configuration is indicative of which type of rearrangement?

  1. A large pericentric inversion
  2. A balanced reciprocal translocation (correct answer)
  3. A Robertsonian translocation
  4. A large interstitial duplication

Explanation: A cross-shaped quadrivalent is the characteristic pairing configuration formed during meiosis in an individual heterozygous for a balanced reciprocal translocation. This structure allows for the alignment of all homologous regions across the two pairs of chromosomes involved. Inversions (A) and duplications (D) form loops, while a Robertsonian translocation (C) typically forms a trivalent. The semi-sterility is explained by the production of unbalanced gametes through adjacent segregation from this quadrivalent.

Question 16

In a paracentric inversion heterozygote, the resolution of a dicentric bridge formed during anaphase I most directly leads to which of the following outcomes?

  1. Nondisjunction of the entire homologous chromosome pair.
  2. The formation of an acentric fragment that is subsequently lost.
  3. Failure of cytokinesis and the creation of a tetraploid cell.
  4. Random breakage of the chromatid, resulting in terminal deletions. (correct answer)

Explanation: A dicentric bridge is formed when the two centromeres of a dicentric chromatid are pulled to opposite poles during anaphase I. This tension causes the chromatid to stretch and eventually break at a random point. This breakage resolves the bridge but produces two chromatids that are missing their terminal ends (terminal deletions) and are thus genetically unbalanced. While the acentric fragment (B) is also formed from the crossover event, it is a separate product, not a consequence of the bridge's resolution. Nondisjunction (A) and failed cytokinesis (C) are different types of meiotic errors.

Question 17

For a carrier of a balanced reciprocal translocation, meiotic segregation of the four chromosomes in the quadrivalent determines the genetic content of the gametes. Which segregation pattern is responsible for producing genetically unbalanced gametes containing one normal and one translocated chromosome?

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

Explanation: In adjacent-1 segregation, homologous centromeres separate at anaphase I, but each pole receives one normal and one translocated chromosome. This results in gametes that are unbalanced, with a duplication of some genetic material and a deletion of other material. In contrast, alternate segregation (A) sends both normal chromosomes to one pole and both translocated chromosomes to the other, producing balanced gametes (either fully normal or balanced carrier). Somatic segregation (C) is not a standard meiotic term, and segregation is not random (D).

Question 18

In Drosophila, an inversion moves the wild-type white gene (w⁺) from a euchromatic region to a new location adjacent to centromeric heterochromatin. This results in flies with mottled eyes, containing both red and white patches. This variable gene expression due to a change in chromosomal position is a classic example of:

  1. Pseudodominance
  2. Haploinsufficiency
  3. Dosage compensation
  4. Position effect variegation (correct answer)

Explanation: Position effect variegation (PEV) occurs when a gene's expression is altered by its new chromosomal location, typically when moved from euchromatin (transcriptionally active) to heterochromatin (transcriptionally repressed). The heterochromatic state can spread into the relocated gene, silencing it in some cells but not others, leading to a variegated or mosaic phenotype. This is distinct from pseudodominance (unmasking a recessive allele), haploinsufficiency (a dosage effect from deletion), and dosage compensation (equalizing gene expression between sexes).

Question 19

A patient with a history of recurrent miscarriages is found to be a carrier of a large chromosomal rearrangement. Analysis of their gametes shows that a significant proportion contain chromosomes that are duplicated for some terminal loci and deleted for other terminal loci. A crossover within the rearranged segment is known to be the cause. Which rearrangement is the most likely diagnosis?

  1. A large paracentric inversion
  2. A balanced reciprocal translocation
  3. A Robertsonian translocation
  4. A large pericentric inversion (correct answer)

Explanation: This specific outcome—gametes containing a chromosome with both a duplication and a deletion of terminal segments—is the hallmark of a crossover event within the inversion loop of a pericentric inversion heterozygote. The crossover produces recombinant chromatids where the segments distal to the inversion are swapped onto chromatids with inverted medial segments, creating the dup/del imbalance. Crossovers in paracentric inversions (A) lead to inviable dicentric/acentric products. Translocations (B, C) produce unbalanced gametes via segregation, which are monosomic or trisomic for large chromosome segments, not typically a single chromosome with both a dup and a del.

Question 20

A researcher identifies a fruit fly with a new phenotype. Genetic mapping reveals that the order of linked genes on chromosome 2, normally A-B-C-D-E, is now A-C-B-D-E in the mutant. This type of rearrangement is best described as:

  1. An interstitial deletion of gene B, followed by reinsertion at a new locus.
  2. A tandem duplication of the B-C region.
  3. A non-reciprocal translocation of gene B.
  4. An interstitial inversion of the B-C segment. (correct answer)

Explanation: The gene order has changed from A-B-C-D-E to A-C-B-D-E. The segment containing genes B and C has been reversed relative to its original orientation. This is the definition of an inversion. Since it is within the arm of a chromosome and does not involve moving to a new chromosome or losing/gaining material, it is an interstitial inversion. The other options describe different rearrangements: deletion/insertion (A), duplication (B), and translocation (C), none of which match the observed change in gene order.