Question 1

Chromosomal aberrations — The exchange of segments between non-homologous chromosomes is best known as what?

Accepted Answer

Introduction / Context: Chromosomal structural changes can profoundly affect gene expression, fertility, and evolution. Distinguishing among crossing over, recombination, and translocation is core genetics knowledge and has practical implications in plant breeding and medical cytogenetics. Given Data / Assumptions: “Non-homologous chromosomes” means chromosomes that do not share the same gene order. Segment exchange between non-homologs alters linkage groups and can create novel gene juxtapositions. Standard meiotic crossing over occurs between homologs, not non-homologs. Concept / Approach: Translocation refers to the movement of a chromosomal segment to a non-homologous chromosome. It can be reciprocal (mutual exchange) or non-reciprocal (one-way transfer). In contrast, recombination is a broad term largely applied to exchange between homologous chromatids during meiosis (crossing over) or to molecular DNA exchange events. Step-by-Step Solution: Identify the actors → non-homologous chromosomes.Match definition → “translocation.”Exclude “crossing over” because it is between homologs. Verification / Alternative check: Classical cytogenetics (e.g., wheat–rye translocations) demonstrates non-homologous exchanges used in breeding for disease resistance. Why Other Options Are Wrong: a) “Translation” is a molecular biology term for protein synthesis.c) Recombination is general; question is specific to non-homologs.d) Crossing over is homologous exchange.e) Endoreduplication duplicates genome without cell division; unrelated. Common Pitfalls: Equating any exchange with “crossing over.” Context (homologous vs non-homologous) is decisive. Final Answer: Translocation.

Question 2

Chromosome number constancy — Within an organism, which process ensures that daughter cells maintain the same chromosome number after cell division?

Accepted Answer

Introduction / Context: Stable inheritance requires that cells maintain the correct chromosome number after each division. The fundamental mechanism is duplication of the entire genome before segregation, ensuring each daughter cell receives a complete set. This concept underlies mitosis in somatic tissues and the orderly reduction in meiosis. Given Data / Assumptions: Cell cycle includes S phase, in which DNA is replicated. Mitosis segregates sister chromatids equally to daughter cells. Meiosis halves chromosome number in gametes; fertilization restores the diploid number across generations. Concept / Approach: Chromosome number constancy from one cell generation to the next relies on accurate replication prior to division. While independent assortment and crossing over generate genetic diversity during meiosis, they do not ensure numerical constancy within somatic divisions. Synapsis is a meiotic pairing event, not the cause of constant number in daughter somatic cells. Step-by-Step Solution: Identify process directly guaranteeing chromatid copies → DNA replication (S phase).Recognize mitotic segregation distributes replicated chromatids equally.Select replication as the key step ensuring same chromosome number post-division. Verification / Alternative check: Cytological observations show chromosomes replicate to form sister chromatids prior to metaphase; aneuploidy arises when replication or segregation fails. Why Other Options Are Wrong: a,b,d) Meiotic processes promote diversity and pairing, not number constancy in somatic divisions.e) Fertilization restores ploidy at organismal level, not cell-to-cell constancy in tissues. Common Pitfalls: Mixing the concept of organismal ploidy restoration (meiosis + fertilization) with cellular chromosome maintenance during mitosis. Final Answer: DNA duplication (replication) prior to cell division.

Question 3

Model organisms — How many chromosomes are present in the budding yeast Saccharomyces cerevisiae (haploid set)?

Accepted Answer

Introduction / Context: Saccharomyces cerevisiae is a premier eukaryotic model organism for genetics, genomics, and cell biology. Its compact genome, excellent genetics, and well-annotated chromosomes make it ideal for studying fundamental processes conserved in higher eukaryotes. Given Data / Assumptions: Yeast exists in haploid and diploid states. Reference genome assemblies identify a fixed chromosome count. Chromosome number questions refer to the standard laboratory strains. Concept / Approach: The canonical haploid chromosome number of S. cerevisiae is 16. Diploids therefore contain 32 chromosomes. Although aneuploidy can occur in laboratory evolution or stress experiments, the baseline genetic map and classical linkage groups are built on 16 chromosomes. Step-by-Step Solution: Recall standard haploid karyotype → 16 chromosomes.Relate to diploid state → 2n = 32.Choose 16 from the options. Verification / Alternative check: Genome databases and textbooks list chromosomes I–XVI, confirming the haploid set totals sixteen. Why Other Options Are Wrong: b,c,d,e) Do not match the established karyotype; they would imply different linkage group counts than observed. Common Pitfalls: Confusing budding yeast with fission yeast (Schizosaccharomyces pombe), which has a different chromosome number. Final Answer: 16.

Question 4

Genome mapping — A chromosome map that uses cloned and sequenced DNA fragments (e.g., restriction fragments, contigs, sequence-tagged sites) rather than classical genetic markers is known as what?

Accepted Answer

Introduction / Context: Genome maps come in multiple types. Genetic (linkage) maps order markers based on recombination frequencies in crosses. Physical maps place DNA features according to actual base-pair distances along chromosomes using molecular tools such as restriction mapping, BAC contigs, and sequence assemblies. Understanding the distinction is vital for positional cloning and genome assembly. Given Data / Assumptions: Question emphasizes cloned and sequenced DNA fragments. Focus is on absolute order and distances in base pairs, not recombination frequency. Terminology follows standard genomics usage. Concept / Approach: A physical map represents the literal arrangement of DNA sequences. Early physical maps used restriction fragment maps and overlapping clone contigs; modern approaches rely on long-read sequencing and scaffolding to create near-complete assemblies. Genetic maps complement physical maps but can differ due to recombination hot/cold spots. Step-by-Step Solution: Identify the map type that uses cloned/sequenced fragments → physical map.Differentiate from genetic map (recombination-based).Exclude vague or nonstandard terms like “cloning map.” Verification / Alternative check: Human genome project milestones included both a high-resolution physical map of BAC contigs and genetic linkage maps; both were necessary but distinct. Why Other Options Are Wrong: b) Genetic maps use centiMorgan distances based on recombination, not cloned fragment positions.c) “Cloning map” is not a standard map category in genomics.d) A correct option exists.e) Epigenetic maps profile chromatin marks or methylation, not cloned fragment positions. Common Pitfalls: Assuming higher recombination frequency equals larger physical distance; recombination landscapes vary significantly across genomes. Final Answer: A physical map.

Question 5

Classical Drosophila genetics—cause of the ‘‘hairy wing’’ phenotype In fruit fly (Drosophila) genetics, the characteristic ‘‘hairy wing’’ trait is primarily caused by which chromosomal change?

Accepted Answer

Introduction / Context: The ‘‘hairy wing’’ phenotype in Drosophila melanogaster is a classical example used in cytogenetics to illustrate how structural chromosome changes can alter gene dosage and produce visible traits. Understanding whether a phenotype results from duplication, deletion, or other rearrangements is a core learning goal in genetics. Given Data / Assumptions: The phenotype involves excessive or ectopic trichomes (hairs) on the wing. The trait is historically associated with altered gene dosage rather than complete loss of function. We are comparing types of chromosome changes: duplication, deletion, aneuploidy (extra X), or point mutation. Concept / Approach: Gene duplications increase copy number of one or more genes. In Drosophila, several classic phenotypes (e.g., Bar eye, hairy wing) arose from segmental duplications that elevate expression of developmental genes, causing patterning changes. Deletions typically reduce structure or viability, extra sex chromosomes cause global sex-dosage effects, and point mutations often alter single-gene function without the characteristic cytological signature of a segmental dosage increase. Step-by-Step Solution: Identify that the phenotype reflects increased expression of loci controlling wing trichomes.Link increased expression to increased gene dosage, a hallmark of duplication.Exclude alternatives: deletion (dose loss), aneuploidy (whole-chromosome sex-dosage issues), simple point mutation (not the classic cytological cause here). Verification / Alternative check: Classical mapping and cytological studies in Drosophila associate hairy wing with duplicated chromosomal segments; restoring normal copy number suppresses the phenotype, consistent with a dosage mechanism. Why Other Options Are Wrong: Deletion: typically reduces structures or causes lethality; does not classically produce hairy wing. Additional X chromosome: affects sex determination and viability, not specifically wing trichome pattern. Point mutation only: possible in principle, but the textbook association for hairy wing is a duplication event. Pericentric inversion: reorders segments but does not inherently change dosage. Common Pitfalls: Confusing any visible trait with a point mutation. Many classic Drosophila markers reflect copy-number changes rather than simple base substitutions. Final Answer: Duplication in a region of a chromosome

Question 6

Chromosome representation in cytogenetics A diagrammatic depiction of the haploid set of chromosomes of an organism, ordered by decreasing size (and banding), is called what?

Accepted Answer

Introduction / Context: Cytogeneticists use standardized visual representations of chromosomes to describe number, size, and banding. Two closely related terms—karyogram and idiogram—are often confused. Clarifying the distinction helps students read primary literature and clinical reports correctly. Given Data / Assumptions: The question specifies a diagram of a haploid set, ordered by decreasing size. Focus is on schematic, not a photographic plate. Banding and centromere positions are shown abstractly in many cases. Concept / Approach: An idiogram (also called ideogram) is a schematic drawing of chromosomes aligned and ordered, typically by size and centromere position, often with banding patterns indicated. A karyogram is the actual arranged image (photograph/micrograph) of metaphase chromosomes. Zymograms relate to enzymes/proteins separated by electrophoresis, not chromosomes. Step-by-Step Solution: Identify that the prompt requests a diagram (not a photo).Recall definitions: idiogram = schematic; karyogram = photographic arrangement.Choose idiogram as the correct descriptor. Verification / Alternative check: Many cytogenetics textbooks define idiogram/ideogram as the diagrammatic representation, often paired with a karyotype description summarizing the chromosome set. Why Other Options Are Wrong: Zymogram: enzyme pattern visualization; unrelated to chromosomes. Chromogram: non-standard term in this context. Karyogram: refers to a photographic arrangement rather than a schematic drawing. Proteogram: protein profile, not cytogenetic. Common Pitfalls: Interchanging idiogram and karyogram; always ask whether the source is a schematic or an image. Final Answer: Idiogram

Question 7

Patterns of sex-linked transmission Transmission of a gene from a male parent to his daughter and then to his grandson (through the daughter) is termed what?

Accepted Answer

Introduction / Context: Sex-linked traits often display characteristic transmission patterns across generations. Recognizing these patterns allows quick identification of X-linked recessive or dominant inheritance in pedigrees. Given Data / Assumptions: Male → daughter → daughter’s son transmission is described. Sex chromosomes: males are XY, females XX. No statement contradicts viability or dominance; the focus is the path of transmission. Concept / Approach: Criss-cross inheritance is typical for X-linked traits: a father passes his X chromosome to all daughters (never to sons). A daughter can then pass that X allele to her sons, producing the “male → daughter → grandson” path. Holandric inheritance is Y-linked (father to all sons), which does not match the pattern here. Step-by-Step Solution: Track the male’s X-linked allele → transmitted to the daughter (who is XX).Daughter passes that X to half of her sons (grandsons of the original male).This alternating sex transmission is termed criss-cross inheritance. Verification / Alternative check: Pedigree analysis examples of hemophilia or color blindness demonstrate this pattern repeatedly. Why Other Options Are Wrong: Holandric: Y-linked goes father → son only. Quantitative: polygenic traits; not a transmission path term. Maternal inheritance: mitochondrial/cytoplasmic, passed from mothers to all offspring. Autosomal dominant: does not impose the sex-specific path described. Common Pitfalls: Assuming all sex-linked traits are father-to-son; X-linked cannot go father-to-son because fathers pass Y to sons. Final Answer: Criss-cross inheritance

Question 8

Dosage compensation in Drosophila melanogaster By what mechanism do male fruit flies balance X-linked gene expression relative to females?

Accepted Answer

Introduction / Context: Different animals solve X-chromosome dosage problems differently. Mammals inactivate one X in females; Drosophila males upregulate their single X. Identifying species-specific strategies is a fundamental comparative genetics concept. Given Data / Assumptions: Males are XY (one X), females XX (two Xs). Dosage compensation equalizes X-linked gene expression between sexes. We are focusing on Drosophila, not mammals. Concept / Approach: In flies, the Male-Specific Lethal (MSL) complex binds the male X chromosome and increases transcription approximately twofold so that male X-linked expression matches that of the two female X chromosomes. Step-by-Step Solution: Recognize species: Drosophila uses upregulation, not X-inactivation.Select the option that states hypertranscription of the male X.Exclude mammalian mechanisms (X-inactivation) and unrelated phenomena. Verification / Alternative check: Genomic and chromatin studies show enrichment of MSL complex and histone acetylation (e.g., H4K16ac) on the male X, consistent with elevated transcription. Why Other Options Are Wrong: Hypotranscription of female Xs is not how Drosophila compensates. X inactivation is mammalian, not Drosophila. Chimerism/imprinting do not explain global dosage compensation in flies. Common Pitfalls: Overgeneralizing mammalian X-inactivation to all animals. Final Answer: Hypertranscription (upregulation) of the male X chromosome

Question 9

Meiotic configurations and structural heterozygosity A cross-shaped (plus-shaped) or ring-of-four configuration at metaphase I typically occurs in which situation?

Accepted Answer

Introduction / Context: Chromosome pairing at meiosis reflects homology. Structural heterozygotes often form distinctive multivalent configurations, which help diagnose rearrangements cytologically. Given Data / Assumptions: Metaphase I of meiosis is considered. Four chromosomes are involved in a stable association. Shape is a cross or ring of four (quadrivalent). Concept / Approach: In reciprocal translocation heterozygotes, parts of nonhomologous chromosomes are exchanged. During meiosis, homologous regions must still pair, generating a quadrivalent—often cross-shaped or ring—so that like segments align. This is distinct from the effects of deletions or simple duplications, which usually cause loops, not a ring-of-four. Step-by-Step Solution: Associate quadrivalents with reciprocal translocations.Exclude deletion (unpaired loops), haploid (univalents, not quadrivalents), and pure duplication (looping of the duplicated segment).Select translocation heterozygote as correct. Verification / Alternative check: Genetic segregation patterns (alternate vs adjacent segregation) and cytology corroborate translocation heterozygosity when quadrivalents are observed. Why Other Options Are Wrong: Deficiency: pairing loop due to missing segment, not ring-of-four. Haploid: typically univalents; quadrivalents require homologous pairs. Duplication: loop of the duplicated segment during synapsis, not a four-chromosome ring. Common Pitfalls: Confusing inversion loops with translocation quadrivalents; inversions produce loops within a single bivalent. Final Answer: Translocation heterozygote (reciprocal translocation)

Question 10

Polyploidy in crop evolution—Raphanobrassica Raphanobrassica (radish × cabbage amphidiploid) is a classical example of which chromosomal condition?

Accepted Answer

Introduction / Context: Allopolyploids arise from hybridization between different species followed by chromosome doubling, restoring fertility. Raphanobrassica is a well-known experimental amphidiploid produced from radish (Raphanus) and cabbage (Brassica). Given Data / Assumptions: Parents belong to different genera/species. Hybrid fertility is restored after chromosome doubling. Resulting organism carries complete chromosome sets from each parent. Concept / Approach: Allopolyploidy (amphidiploidy) involves combining two distinct but complete genomes. This is distinct from aneuploidy (loss/gain of one or few chromosomes) and nullisomy (loss of a homologous pair). Autopolyploidy duplicates the same genome, not different ones. Step-by-Step Solution: Recognize radish × cabbage hybrid as interspecific.Chromosome doubling yields an amphidiploid with pairing partners restored.Conclude allopolyploidy as the correct category. Verification / Alternative check: Cytological pairing and fertility recovery after doubling confirms amphidiploid status in such hybrids. Why Other Options Are Wrong: Aneuploidy/nullisomy: partial chromosome number changes, not whole-genome combinations. Autopolyploidy: duplicates the same genome rather than merging two species’ genomes. Haploidy: half the somatic chromosome number, typically sterile. Common Pitfalls: Confusing any polyploid with allopolyploid; always ask if the genomes are from different species. Final Answer: Allopolyploidy (amphidiploidy)

Question 11

Phenotypic outcome of a genotype—key definitions The degree to which a given genotype is manifested in the phenotype of an individual is termed what?

Accepted Answer

Introduction / Context: Two related terms—penetrance and expressivity—describe different aspects of how genotypes translate into phenotypes. Misunderstanding these terms leads to errors in interpreting pedigrees and risk assessments. Given Data / Assumptions: We focus on the extent or degree of trait expression in an individual who carries the genotype. We are not being asked what fraction of carriers show the trait. Concept / Approach: Expressivity measures the severity or range of phenotypic expression among individuals with the same genotype (e.g., mild to severe). Penetrance measures the proportion of individuals with the genotype who actually show any phenotype at all (all-or-none at the individual level). Therefore, “degree or extent” points to expressivity. Step-by-Step Solution: Identify the keyword “degree or extent” → expressivity.Distinguish from penetrance (presence vs absence).Select expressivity as the correct term. Verification / Alternative check: Clinical genetics routinely documents variable expressivity (e.g., neurofibromatosis) where all carriers may show signs, but severity varies widely. Why Other Options Are Wrong: Penetrance: fraction expressing the trait, not the degree of expression. Inheritance strength/environmental influence/additive variance: broader concepts, not the specific definition asked. Common Pitfalls: Using “penetrance” whenever variation is mentioned; remember penetrance is yes/no, expressivity is how much. Final Answer: Expressivity

Question 12

Quantitative traits and continuous variation When a character displays continuous variation with an approximately normal distribution in a population, which statement best explains the genetic basis?

Accepted Answer

Introduction / Context: Traits such as height, yield, or skin color often vary continuously and approximate a bell-shaped curve. These quantitative traits reflect polygenic inheritance and environmental influences. Given Data / Assumptions: Distribution is continuous and near-normal. No single discrete classes dominate the phenotype. Environmental effects may contribute but do not fully explain the pattern. Concept / Approach: Polygenic traits result from many loci, each with small additive effects. According to the central limit tendency, the sum of many small, independent genetic contributions plus environmental noise yields a normal-like distribution, consistent with observed quantitative traits. Step-by-Step Solution: Recognize continuous, near-normal distribution → polygenic control.Acknowledge that environment modulates variance but is not the sole cause.Choose the polygenic explanation as best fit. Verification / Alternative check: Biometric analyses (heritability, QTL mapping) consistently reveal multiple loci with small effects underlying quantitative traits. Why Other Options Are Wrong: “Not controlled genetically”: contradicts QTL evidence. “Climate only”: environment contributes but cannot alone produce consistent heritable distribution. Single-gene complete dominance: would yield discrete classes, not smooth normal curves. Common Pitfalls: Attributing all variation to environment; overlooking additive genetic variance. Final Answer: It is governed by many genes with small cumulative effects (polygenic)

Question 13

Expected phenotypic ratios in Mendelian dihybrid crosses If an F1 individual is heterozygous for two independently assorting genes and is selfed (or mated inter se with a like F1), what phenotypic ratio is expected in the F2?

Accepted Answer

Introduction / Context: The hallmark ratio for a Mendelian dihybrid cross under independent assortment is 9:3:3:1 in the F2 generation. This assumes complete dominance at each locus, no epistasis, and random segregation. Given Data / Assumptions: One F1 is heterozygous at both loci (AaBb). Cross is an F1 self (AaBb × AaBb) or inter se with another AaBb. Genes assort independently; dominance is complete at each locus. Concept / Approach: Each locus segregates 3:1 for the dominant phenotype. The joint probability across two independent loci yields the product distribution: 9 double-dominant phenotypes, 3 dominant at A only, 3 dominant at B only, and 1 double-recessive phenotype. Step-by-Step Solution: Compute A locus: 3 dominant : 1 recessive.Compute B locus: 3 dominant : 1 recessive.Combine: (33)=9 both dominant; (31)=3 A dominant only; (13)=3 B dominant only; (11)=1 both recessive → 9:3:3:1. Verification / Alternative check: Punnett square or probability tree confirms the same ratio. A testcross (AaBb × aabb) would instead yield 1:1:1:1, highlighting the importance of the specified mating. Why Other Options Are Wrong: 1:1:1:1: testcross outcome, not F1 self. Other ratios do not follow independent assortment with complete dominance. Common Pitfalls: Confusing F1 self with testcross; always check the mating design. Final Answer: 9:3:3:1

Question 14

Cytogenetic effects of diethylstilbestrol (DES) Which cellular effect is most characteristically associated with DES exposure during cell division studies?

Accepted Answer

Introduction / Context: Diethylstilbestrol (DES) is a synthetic estrogen historically used in medicine and later recognized for adverse developmental and carcinogenic effects. In cytogenetic assays, certain chemicals act as clastogens (causing chromosome breaks) or aneugens (interfering with the mitotic spindle and chromosome segregation). Distinguishing these mechanisms improves interpretation of micronucleus and karyotypic data. Given Data / Assumptions: Focus is on mitotic effects observable in cell division studies. We compare spindle disruption versus isolated phenomena like centriole elongation. Endpoints such as lagging chromosomes can be downstream consequences of spindle defects. Concept / Approach: Aneugens alter the spindle apparatus or kinetochore function, promoting mis-segregation (nondisjunction, lagging chromatids). DES has been reported in several test systems to perturb microtubule dynamics and spindle function, consistent with aneugenicity. While chromosome lagging may be observed, it is typically a consequence of spindle interference rather than the fundamental, defining action itself. Step-by-Step Solution: Identify DES as a chemical with endocrine activity and mitotic effects.Associate observed lagging chromosomes with underlying spindle disruption.Select spindle disruption (aneugen) as the primary mechanistic descriptor. Verification / Alternative check: Micronucleus assays distinguishing kinetochore-positive micronuclei (aneugenic) from kinetochore-negative (clastogenic) support spindle interference when kinetochore-positive bodies increase after DES exposure in certain systems. Why Other Options Are Wrong: Chromosome lagging only: a sign, not the root mechanism. Centriole elongation: not the primary, broadly reported cytogenetic effect of DES. All equally: overgeneralizes; evidence points most strongly to spindle interference. No effect: contradicted by numerous toxicology studies. Common Pitfalls: Equating any visible mitotic anomaly with the mechanism; always link the phenotype back to spindle or DNA damage pathways. Final Answer: Disruption of the mitotic spindle (aneugenic effect)

Question 15

In classical genetics, what do we call the percentage of individuals with a specific genotype who actually express the corresponding phenotype (i.e., who show the trait under standard conditions)?

Accepted Answer

Introduction / Context: In human genetics and Mendelian inheritance, the terms “penetrance” and “expressivity” are often confused. This question tests your understanding of how often a genotype produces its expected phenotype in real organisms, which is a cornerstone for risk prediction in medical genetics and for interpreting pedigree data. Given Data / Assumptions: A particular genotype is known (for example, a dominant disease allele). We observe a set of individuals carrying that genotype. We assess whether they show the associated phenotype. Concept / Approach: Penetrance refers to the proportion (often expressed as a percentage) of individuals with a given genotype who display the expected phenotype. If 80 out of 100 carriers show the trait, penetrance is 80%. Expressivity, by contrast, describes the degree or severity of phenotype among those who are affected. Distinguishing these concepts is essential when evaluating variable outcomes from the same genotype. Step-by-Step Solution: Define what is being measured: proportion of genotype-positive individuals who show the trait.Recall that penetrance = (number showing phenotype / total with genotype) * 100%.Recognize that expressivity addresses intensity or range of symptoms, not frequency of being affected.Identify distractors: “inheritance percent” is not a formal term; “environmental influence” can modify penetrance but is not the definition. Verification / Alternative check: In disorders with incomplete penetrance, not all carriers are affected; clinical genetics references consistently define penetrance as this percentage, confirming the term selection. Why Other Options Are Wrong: Expressivity: measures variation in severity among affected individuals, not the proportion affected. Inheritance percent: nonstandard terminology. Environmental influence: a factor that modulates penetrance, not the definition itself. Trait prevalence: frequency of the phenotype in a population regardless of genotype, not conditioned on genotype. Common Pitfalls: Confusing “how many show it” (penetrance) with “how strongly they show it” (expressivity). Also, mixing up population prevalence with genotype-conditioned probability. Final Answer: Penetrance

Question 16

Chromosomal inversions and meiosis: An odd number of crossovers within an inverted segment will yield a dicentric chromatid and an acentric fragment at anaphase I in which kind of inversion?

Accepted Answer

Introduction / Context: Inversions alter the linear order of genes and can produce meiotic complications. Whether the centromere is included in the inverted segment determines the products of crossing over and the viability of resulting gametes. This question probes your ability to link cytogenetic configurations to meiotic outcomes. Given Data / Assumptions: A single crossover occurs within an inverted region during meiosis I. We are comparing paracentric (centromere outside) versus pericentric (centromere inside) inversions. We observe anaphase I chromosome behavior. Concept / Approach: In a paracentric inversion, a crossover within the loop generates one dicentric chromatid (two centromeres) and one acentric fragment (no centromere). The dicentric can form a bridge that breaks; the acentric is lost, often causing gamete inviability. In pericentric inversions, crossovers yield chromatids with duplications and deletions but not dicentric/acentric products because the centromere is included in the exchanged segments. Step-by-Step Solution: Identify inversion type: paracentric excludes the centromere.Model a single crossover inside the looped region.Track centromere number: one recombinant chromatid ends up dicentric; the reciprocal is acentric.Conclude that the described outcome matches paracentric inversion. Verification / Alternative check: Cytology textbooks consistently illustrate paracentric crossovers with bridges and fragments at anaphase I; pericentric events instead create duplication/deletion chromatids that retain single centromeres. Why Other Options Are Wrong: Pericentric inversion: produces duplication/deletion chromatids, not dicentric/acentric pairs. Double inversion: undefined here and does not specifically predict the classic dicentric/acentric outcome. None of these and isochromosome: do not match the hallmark products described. Common Pitfalls: Memorizing outcomes without distinguishing whether the centromere lies inside or outside the inversion loop. Final Answer: Paracentric inversion (excludes the centromere)

Question 17

Model organism fact check: How many chromosomes (haploid count) does the nematode Caenorhabditis elegans have?

Accepted Answer

Introduction / Context: Caenorhabditis elegans is a premier genetic model, with a compact, well-mapped genome. Knowing its chromosome number is foundational for interpreting linkage maps, genome assemblies, and sex determination in this organism. Given Data / Assumptions: We are asked for the chromosome count characteristic of C. elegans. The standard laboratory strain is hermaphroditic (XX) with males (XO) occurring at low frequency. Concept / Approach: C. elegans has six chromosomes in the haploid set: five autosomes (I–V) and one sex chromosome (X). Hermaphrodites are XX (2n = 12), males are XO (2n = 11). The neat, small karyotype supports clear genetic analysis and fully sequenced reference assemblies per chromosome. Step-by-Step Solution: Recall autosomes: I, II, III, IV, V = five autosomes.Add the sex chromosome X to total haploid count.Compute: 5 autosomes + 1 sex chromosome = 6 haploid chromosomes.Select “6”. Verification / Alternative check: Genome browsers and reference papers list the six chromosome scaffolds by name (I–V, X), confirming the haploid number. Why Other Options Are Wrong: 3, 9, 12: do not match known karyotype; 12 would be the diploid number in hermaphrodites, not haploid. 5: omits the X chromosome. Common Pitfalls: Confusing haploid with diploid number, or mixing up autosomes and sex chromosomes. Final Answer: 6

Question 18

Mendelian terminology refresher: Which term denotes a pair of contrasting hereditary factors that occupy the same locus on homologous chromosomes?

Accepted Answer

Introduction / Context: Understanding core genetic vocabulary is vital for solving inheritance problems. Mendel’s “factors” correspond to genes, and the alternative versions that account for contrasting traits (such as tall vs. dwarf) are called alleles. Given Data / Assumptions: Two contrasting forms of a trait are considered (e.g., round vs. wrinkled seeds). They occupy the same locus on homologous chromosomes in a diploid organism. Concept / Approach: Alleles are alternative forms of the same gene. Phenotype is the observed trait. Genotype is the genetic constitution (e.g., AA, Aa, aa). Homozygous refers to identical alleles at a locus. Haplotype refers to a set of alleles on one chromosome segment transmitted together. Only “alleles” specifically denotes contrasting forms at the same locus. Step-by-Step Solution: Identify the locus: a fixed chromosomal position for a gene.Recognize alternate versions of the gene: alleles.Exclude descriptive terms of state (phenotype, genotype, homozygous).Select “Alleles”. Verification / Alternative check: Standard genetics texts define allele as one of two or more alternative forms of a gene at a locus, responsible for contrasting characters. Why Other Options Are Wrong: Phenotype: outward appearance, not the genetic variants themselves. Homozygous: condition of having identical alleles, not the alleles. Genotype: full genetic makeup, more general than alleles at a single locus. Haplotype: linked allele set on one chromosome, not necessarily contrasting forms at one locus. Common Pitfalls: Equating genotype with allele or thinking phenotype refers to genes rather than traits. Final Answer: Alleles

Question 19

Genetic linkage logic: If two genes display 50% recombination in test crosses, what does this imply about their chromosomal relationship?

Accepted Answer

Introduction / Context: Recombination frequency is a proxy for physical distance, but it plateaus at 50%. Understanding this ceiling is essential to interpret linkage data correctly and avoid false conclusions about gene location from genetic crosses alone. Given Data / Assumptions: Measured recombination fraction between two loci is approximately 0.50 (50%). Standard meiotic crossing-over and independent assortment occur. No transmission distortion or mapping anomalies are assumed. Concept / Approach: Recombination cannot exceed 50% in conventional meiotic mapping because at that point gamete classes show independent assortment. Two scenarios yield 50%: (1) genes on different chromosomes; (2) genes on the same chromosome but so far apart that multiple crossovers randomize associations, making them behave as if unlinked. Thus, 50% recombination does not distinguish between these two possibilities. Step-by-Step Solution: Recall the definition: recombination fraction r ranges from 0 to 0.5.Recognize r = 0.5 corresponds to independent assortment.Note two causes: different chromosomes, or very distant loci on the same chromosome.Therefore select the combined option acknowledging both explanations. Verification / Alternative check: Classical linkage maps and population genetics texts describe the 50% ceiling and the indistinguishability of very distant linkage from actual independence using recombination data alone. Why Other Options Are Wrong: Only (a) or only (b) would be incomplete; either can be true. Tight linkage implies recombination far below 50%. Y chromosome location is unrelated to the general principle. Common Pitfalls: Assuming 50% automatically means different chromosomes; remember that long chromosomes can also yield 50% due to multiple crossovers. Final Answer: Both (a) and (b) are valid explanations for 50% recombination.

Question 20

Naming inversion types: When an inverted chromosomal segment includes the centromere, what is this inversion called?

Accepted Answer

Introduction / Context: Inversions are structural rearrangements that reverse a chromosome segment. Correctly naming them based on whether the centromere is within the inverted block is essential for predicting meiotic outcomes and cytogenetic behavior. Given Data / Assumptions: A chromosome harbors an inversion. The inverted segment explicitly includes the centromere. Concept / Approach: A pericentric inversion includes the centromere; a paracentric inversion does not. This distinction determines whether crossovers produce duplication/deletion chromatids (pericentric) or dicentric/acentric products (paracentric). Naming conventions are universal in cytogenetics for clarity in clinical and research contexts. Step-by-Step Solution: Check whether the centromere lies inside the inverted block.If yes, classify as pericentric inversion.If no, classify as paracentric inversion.Select “Pericentric inversion”. Verification / Alternative check: Standard cytogenetic atlases define these terms identically, aligning with observed meiotic segregation patterns. Why Other Options Are Wrong: Paracentric: excludes the centromere. Double/inverted inversion: not standard nomenclature here. Isochromosome: a different abnormality involving mirror-image arms, not an inversion. Common Pitfalls: Confusing the outcomes of crossovers for paracentric versus pericentric inversions and then misnaming the inversion type. Final Answer: Pericentric inversion

Animal Biotechnology Chromosomes Questions