Chromosomal Basis of Inheritance (1C)
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MCAT Biological and Biochemical Foundations of Living Systems › Chromosomal Basis of Inheritance (1C)
In a plant species, a recessive allele d causes dwarfism. A dwarf plant (d/d) is crossed with a heterozygous plant (D/d). Cytological analysis of the heterozygous parent reveals a reciprocal translocation between two nonhomologous chromosomes; the heterozygote is a translocation carrier but phenotypically normal. In the offspring, many seeds abort early, and among the surviving seedlings the ratio of normal:dwarf is close to 1:1. Which explanation best accounts for the observed seed abortion and the surviving phenotype ratio?
Crossing over is eliminated in translocation heterozygotes, forcing all gametes to be parental and causing dwarfism in half the survivors
Adjacent segregation in a translocation heterozygote produces many unbalanced gametes that lead to embryo lethality, while balanced gametes yield an apparent 1:1 D/d to d/d ratio
Nondisjunction in meiosis II creates trisomic embryos that abort, while monosomic embryos survive and appear as dwarf plants
Independent assortment is disrupted by the translocation, causing preferential transmission of the D allele and resulting in seed abortion
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how reciprocal translocations affect meiotic segregation and offspring viability. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but structural rearrangements like translocations create special segregation patterns. In this context, the translocation heterozygote forms a quadrivalent during meiosis, and adjacent segregation produces many unbalanced gametes with duplications/deletions that cause embryo lethality. The correct choice A is consistent because it aligns with translocation heterozygotes producing both balanced gametes (yielding viable 1:1 D/d to d/d offspring) and unbalanced gametes (causing seed abortion). Choice D is incorrect because it misinterprets the abortion as due to nondisjunction creating trisomies, a common error when students assume all chromosome imbalances involve whole chromosome number changes. When analyzing chromosomal inheritance, ensure understanding that translocations create partial imbalances through adjacent segregation and verify predictions by considering which gametes contain balanced chromosome segments.
In a diploid frog species, a specific autosome carries a visible cytological inversion in one homolog (an inversion heterozygote). Researchers cross an inversion heterozygote to a normal homozygote and score offspring viability. They observe a substantial fraction of embryos that arrest early, while surviving offspring inherit either the normal arrangement or the inverted arrangement without obvious recombination within the inverted segment. Which explanation best accounts for reduced viability and suppressed recovery of recombinants within the inversion?
Sister chromatids separate in meiosis I in inversion heterozygotes, preventing recombination and causing embryo lethality
Inversions increase crossing over frequency, so most embryos are recombinant and die due to expression of recessive alleles
Independent assortment fails in inversion heterozygotes, causing embryos to inherit both homologs and arrest due to trisomy
Crossing over within the inversion loop can produce unbalanced chromatids (with duplications/deletions), reducing viability; recombinants are therefore rarely recovered among survivors
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how chromosomal inversions suppress recombination and affect offspring viability. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but inversions create special problems during crossing over. In this context, crossing over within an inversion loop produces dicentric (two centromeres) and acentric (no centromere) chromatids that are lost during division, creating unbalanced gametes with duplications and deletions. The correct choice A is consistent because it aligns with inversion heterozygotes producing lethal imbalances from crossovers, explaining both embryo arrest and suppressed recovery of recombinants among survivors. Choice D is incorrect because it misinterprets inversions as increasing crossing over, a common error when students don't understand that inversions suppress functional recombination recovery. When analyzing chromosomal inheritance, ensure understanding that inversions don't prevent crossing over but make recombinant products inviable, and verify predictions by considering the fate of crossover products within inversion loops.
In a laboratory strain of budding yeast (Saccharomyces cerevisiae), a diploid cell is heterozygous for two loci on the same chromosome: LEU2 (Leu+ vs leu−) and HIS3 (His+ vs his−). The cell is induced to sporulate, and tetrads are dissected. For a subset of tetrads, the four spores show the following viable phenotypes: 2 spores are Leu+ His− and 2 spores are leu− His+. No Leu+ His+ or leu− his− spores are observed in those tetrads. Based on chromosomal behavior during meiosis, which explanation best accounts for this inheritance pattern in that subset of tetrads?
A single crossover between LEU2 and HIS3 producing a tetratype arrangement of alleles
Independent assortment of LEU2 and HIS3 due to their location on different chromosomes
Nondisjunction at meiosis II producing two disomic and two nullisomic spores for the chromosome
No crossover between LEU2 and HIS3, with the diploid in repulsion phase (trans) for the two loci
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how linked genes segregate during meiosis. Chromosomal behaviors such as segregation and crossing over during meiosis determine the distribution of alleles in gametes. In this context, the observation of only Leu+ His- and leu- His+ spores indicates that the two loci are linked on the same chromosome and the diploid parent had these alleles in repulsion phase (trans configuration). The correct choice is consistent because when no crossover occurs between linked loci in trans configuration, only parental-type gametes are produced. Choice B is incorrect because a single crossover would produce all four possible phenotypes (tetratype), not just two parental types. When analyzing chromosomal inheritance, verify that the observed gamete classes match the expected outcomes based on the initial allele arrangement and presence or absence of recombination.
In a nematode (C. elegans) strain, a researcher follows a fluorescent tag integrated near the centromere of chromosome V. During oogenesis, microscopy shows that in some oocytes, both homologs of chromosome V move to the same pole at anaphase I. The resulting embryos frequently arrest early in development. Which outcome is most consistent with chromosomal segregation during meiosis in those oocytes?
Gametes produced from those oocytes will be aneuploid for chromosome V, leading to monosomic or trisomic embryos after fertilization
Gametes will show increased recombination near the centromere, preventing aneuploidy
Gametes will be genetically normal because sister chromatids, not homologs, separate at anaphase I
All embryos should be viable because meiosis II can correct homolog mis-segregation from meiosis I
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how meiosis I nondisjunction produces aneuploid gametes. Chromosomal behaviors during anaphase I normally separate homologous chromosomes to opposite poles, ensuring balanced chromosome numbers. In this context, both chromosome V homologs moving to the same pole creates gametes with either two copies or zero copies of chromosome V. The correct choice is consistent because these aneuploid eggs will produce trisomic (three copies) or monosomic (one copy) embryos after fertilization with normal sperm. Choice A is incorrect because meiosis II cannot correct homolog mis-segregation from meiosis I - the damage is already done. When analyzing meiotic errors, trace chromosome movements through both divisions to predict the final gamete composition and resulting embryo viability.
In a mammalian cell line, investigators induce a deletion on one homolog of chromosome 5 that removes a gene required for normal craniofacial development. The other homolog carries a functional allele. Cells with the deletion show reduced gene product and a measurable craniofacial phenotype in a mouse model generated from these cells. Which chromosomal explanation best accounts for a phenotype arising despite the presence of one functional allele?
Segregation of sister chromatids in meiosis I, which duplicates the deletion onto the intact homolog
Haploinsufficiency due to a chromosomal deletion, where one functional copy does not produce enough gene product for a normal phenotype
Crossing over between homologs in mitosis, which converts the functional allele into a nonfunctional allele in all tissues
Independent assortment of chromosome 5 during meiosis, which prevents the functional allele from being inherited
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how chromosomal deletions can cause haploinsufficiency phenotypes. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but structural changes like deletions can reveal dosage-sensitive genes. In this context, the deletion removes one copy of a gene required for craniofacial development, and the single remaining functional copy cannot produce sufficient gene product for normal development. The correct choice D is consistent because it aligns with haploinsufficiency, where one functional allele is insufficient for wild-type phenotype in dosage-sensitive genes. Choice C is incorrect because it misinterprets the mechanism as crossing over in mitosis converting alleles, a common error when students confuse germline and somatic events. When analyzing chromosomal inheritance, ensure understanding that some genes require two functional copies for normal phenotype and verify predictions by considering gene dosage effects in deletion heterozygotes.
A plant geneticist analyzes a diploid maize line heterozygous for two markers on chromosome 3: A/a and B/b. The line is testcrossed to aabb. Among 200 offspring, 92 show phenotype A B, 88 show a b, 10 show A b, and 10 show a B. No other phenotypes are observed. Which explanation best accounts for the inheritance pattern observed?
Genes A and B are linked on the same chromosome, and most offspring reflect parental (nonrecombinant) gametes
Meiosis II nondisjunction at chromosome 3 produces two recombinant classes at high frequency
A dominant lethal allele eliminates recombinant zygotes, so recombination cannot occur between A and B
Genes A and B assort independently because they are on different chromosomes, yielding a 1:1:1:1 distribution
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how linkage affects segregation ratios in testcrosses. Chromosomal behaviors such as independent assortment produce equal frequencies of all gamete types, while linkage reduces recombinant frequencies. In this context, the high frequency of AB (92) and ab (88) phenotypes compared to Ab (10) and aB (10) indicates these loci are linked. The correct choice is consistent because linked genes on the same chromosome produce mostly parental-type gametes, with recombinants arising only from crossing over events. Choice B is incorrect because independent assortment would yield approximately 50 of each phenotypic class, not the observed 92:88:10:10 ratio. When analyzing inheritance patterns, calculate recombination frequency (20/200 = 10%) to confirm linkage and estimate map distance between loci.
A clinical genetics lab analyzes karyotypes from embryos of a mouse model. A subset shows trisomy of chromosome 16 (three copies), while the remainder are euploid. In oocytes from the trisomy-positive group’s mothers, microscopy during meiosis reveals occasional failure of homologous chromosomes to separate during anaphase I.
Based on the chromosomal behavior described, what is most likely the chromosomal composition of the resulting abnormal zygote immediately after fertilization by a normal haploid sperm?
Trisomy 16 because the egg received two homologs of chromosome 16 and the sperm contributed one
Triploidy for all chromosomes because nondisjunction affects the entire genome
Monosomy 16 because both homologs moved to the same pole at anaphase II
Normal disomy 16 because crossing over corrects nondisjunction at anaphase I
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how nondisjunction during meiosis leads to aneuploidy. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but errors like nondisjunction disrupt chromosome number. In this context, the failure of homologs to separate at anaphase I in mouse oocytes illustrates how this produces disomic gametes. The correct choice is consistent because it aligns with the egg receiving two homologs, leading to trisomy upon fertilization. Choice A is incorrect because it misinterprets the timing of nondisjunction, a common error when students confuse meiosis I and II divisions. When analyzing chromosomal inheritance, ensure understanding of basic meiotic principles and verify predictions with known genetic laws.
In a human pedigree study, an autosomal dominant trait appears in every generation. However, a genotyped affected parent (heterozygous) has one child who lacks the trait and is confirmed to have inherited the parent’s chromosome carrying the normal allele at the trait locus. No evidence of mutation is found. The trait locus is near a polymorphic marker used for haplotyping, and a crossover is detected between the marker and the trait locus in the meiosis that produced the unaffected child.
Which explanation best accounts for this observation?
Independent assortment can occur between loci on the same chromosome only if they are dominant
Segregation of alleles fails when a trait is autosomal dominant, causing apparent recombination
Crossing over can separate a trait allele from a nearby marker, changing which haplotype co-segregates with the phenotype
Nondisjunction at meiosis II produces a normal child by eliminating the dominant allele
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how recombination can alter haplotype transmission. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity through crossing over. In this context, the crossover detected between the marker and trait locus illustrates recombination separating alleles. The correct choice is consistent because it aligns with crossing over changing the inherited haplotype. Choice C is incorrect because it misinterprets segregation in dominant traits, a common error when assuming dominance affects meiotic mechanics. When analyzing chromosomal inheritance, ensure understanding of basic meiotic principles and verify predictions with known genetic laws.
In a frog species, sex is determined by ZW females and ZZ males. A female with genotype ZW undergoes meiosis, but nondisjunction of the sex chromosomes occurs at anaphase I in a subset of oocytes. She mates with a normal ZZ male.
Which zygotic sex chromosome complement is most consistent with an egg produced by meiosis I nondisjunction and fertilized by a normal sperm?
ZZ
ZW
WW
ZZW
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how nondisjunction alters sex chromosome complements. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but errors produce aneuploids. In this context, meiosis I nondisjunction in the frog oocyte illustrates disomic eggs. The correct choice is consistent because it aligns with ZZW trisomy from disomic egg and Z sperm. Choice C is incorrect because it misinterprets nondisjunction outcomes, a common error when confusing ZW segregation. When analyzing chromosomal inheritance, ensure understanding of basic meiotic principles and verify predictions with known genetic laws.
In a plant species used for breeding experiments, two genes controlling seed traits (R/r for red vs white seed coat; S/s for smooth vs wrinkled surface) are located on the same chromosome. A heterozygous plant in coupling phase (RS/rs) is testcrossed to a homozygous recessive plant (rs/rs). Progeny show mostly RS and rs phenotypes, with fewer Rs and rS.
Which explanation best accounts for the observed progeny distribution?
Sister chromatids segregate at meiosis I, producing only parental gametes in the absence of nondisjunction
Independent assortment requires genes to be on the same chromosome in order to segregate randomly
Linked genes tend to be inherited together because recombination between them is less frequent than segregation of homologs
Dominant alleles suppress crossing over, reducing recombinant classes
Explanation
This question assesses understanding of the chromosomal basis of inheritance, specifically how linkage influences progeny distributions. Chromosomal behaviors such as segregation and independent assortment during meiosis ensure genetic diversity, but linked genes deviate from 1:1:1:1 ratios. In this context, the genes on the same chromosome in the plant illustrate reduced recombination. The correct choice is consistent because it aligns with linkage causing more parental than recombinant progeny. Choice B is incorrect because it misinterprets linkage as requiring same-chromosome placement for random segregation, a common error when confusing linkage with assortment. When analyzing chromosomal inheritance, ensure understanding of basic meiotic principles and verify predictions with known genetic laws. Use testcross data to map gene distances.