Gamete Outcomes from Meiosis - Genetics

If one ascus contains the spore arrangement (arg-, arg-, arg-, arg-, arg+, arg+, arg+, arg+), what meiotic event does this pattern represent?
Answer: A first division segregation pattern, indicating no crossover occurred between the arg gene and the centromere. In an ordered octad, the arrangement of spores reflects meiotic segregation. A 4:4 pattern (four spores of one type followed by four of the other) indicates that the alleles segregated during meiosis I. This is known as first division segregation (FDS) and occurs when there is no crossover between the gene locus and the centromere. A second division segregation (SDS) pattern, such as 2:4:2 or 2:2:2:2, would indicate a crossover occurred in that region.
A primary spermatocyte has a pair of homologous chromosomes where one member has a terminal deletion that removes locus B. Its homolog is normal and carries allele B. This cell is also heterozygous for an unlinked gene C (genotype Cc). After a normal meiosis, what proportion of the sperm are expected to have the normal B allele and the c allele?
Answer: 1/4 Segregation of the B locus and the C locus can be considered independently. For the B locus, the two homologous chromosomes are one with allele B and one with a deletion (let's denote it '-'). After meiosis, half the sperm will receive the chromosome with B, and half will receive the chromosome with the deletion. So, P(B) = 1/2. For the unlinked C locus, the genotype is Cc. Half the sperm will receive C, and half will receive c. So, P(c) = 1/2. The probability of a sperm containing both B and c is the product of their individual probabilities: P(B and c) = P(B) × P(c) = 1/2 × 1/2 = 1/4.
An individual is heterozygous for a large pericentric inversion on chromosome 3. A single crossover event occurs within the inversion loop during meiosis. Which of the following is an accurate description of the gametes produced from this single meiotic event?
Answer: Two non-recombinant gametes (one normal, one inverted) and two non-viable recombinant gametes, each with a duplication and a deletion. A crossover within the loop of a pericentric inversion (which includes the centromere) generates four distinct chromatids. Two are the original non-recombinant chromatids (one with the normal sequence, one with the inverted sequence). The other two are recombinant chromatids that are genetically unbalanced; each contains a duplication of the segment outside the inversion on one end and a deletion of the segment on the other end. Gametes receiving these unbalanced chromatids are typically non-viable. A crossover in a paracentric inversion would produce dicentric and acentric chromosomes.
Genes A and B are completely linked on an autosome. A male with genotype AB/ab undergoes spermatogenesis. In one primary spermatocyte, nondisjunction of this chromosome occurs during meiosis II in the secondary spermatocyte that received the AB chromosome. What distinct genotypes of sperm will be produced from this single meiotic event?
Answer: Sperm with genotypes ab, AB/AB, and nullisomic for the chromosome. Meiosis I separates the homologous chromosomes, so one secondary spermatocyte receives the AB chromosome and the other receives the ab chromosome. The 'ab' cell undergoes a normal meiosis II, producing two 'ab' sperm. The 'AB' cell undergoes nondisjunction in meiosis II, meaning its sister chromatids (both AB) fail to separate. This yields one diploid (n+1) sperm with two copies of the AB chromosome (genotype AB/AB) and one nullisomic (n-1) sperm lacking this chromosome entirely. Thus, the four products from the original cell are two 'ab' sperm, one 'AB/AB' sperm, and one nullisomic sperm.
In Drosophila, gray body (b+) is dominant to black body (b), and normal wings (vg+) are dominant to vestigial wings (vg). These two genes are linked. A female fly of genotype b+ vg+ / b vg undergoes oogenesis, but nondisjunction of this chromosome pair occurs during meiosis I. If the resulting n+1 egg is fertilized by a sperm from a black-bodied, vestigial-winged male, what is the expected phenotype of the resulting zygote?
Answer: Gray body, normal wings Nondisjunction in meiosis I means the homologous chromosomes fail to separate. The female (b+ vg+ / b vg) will produce an n+1 egg containing both homologous chromosomes, so its genotype is b+ vg+ / b vg. The sperm from a black, vestigial male (b vg / b vg) has the genotype b vg. Fertilization results in a zygote with the genotype b+ vg+ / b vg / b vg. Since b+ is dominant to b, and vg+ is dominant to vg, the presence of the b+ and vg+ alleles will result in a wild-type phenotype: gray body and normal wings.
An oocyte from a female with genotype X^AX^a undergoes meiosis. A nondisjunction event occurs. Which of the following sets of genotypes for the mature ovum and its corresponding polar bodies is only possible if the nondisjunction event occurred during meiosis II?
Answer: Ovum: X^AX^A; First polar body: X^a; Second polar body: nullisomic A nondisjunction event in meiosis II involves the failure of sister chromatids to separate. If meiosis I proceeds normally, the secondary oocyte will receive either X^A or X^a, and the first polar body will receive the other. If the secondary oocyte containing X^A undergoes nondisjunction, its sister chromatids (both X^A) will both move to the ovum, resulting in an X^AX^A ovum and a nullisomic second polar body. The first polar body (X^a) would divide normally, though it typically degenerates. An ovum genotype of X^AX^a can only result from nondisjunction of homologous chromosomes in meiosis I.
An individual is heterozygous for a paracentric inversion. A primary spermatocyte from this individual undergoes a single crossover within the inversion loop. Which of the following describes the complete set of four products at the conclusion of meiosis II?
Answer: One viable gamete with a normal chromosome, one viable gamete with the inverted chromosome, and two non-viable gametes with deletions. A crossover within a paracentric inversion loop creates a dicentric chromatid (two centromeres) and an acentric fragment (no centromere). During anaphase I, the dicentric chromatid forms a bridge that eventually breaks, while the acentric fragment is lost. This results in two of the four final gametes receiving chromosomes with massive deletions, rendering them non-viable. The other two chromatids were not involved in the crossover and remain intact: one is the normal parental chromosome and one is the inverted parental chromosome. These segregate to produce two viable gametes.
A male with Klinefelter syndrome (47,XXY) undergoes spermatogenesis. During meiosis I, the two X chromosomes form a bivalent and the Y chromosome acts as a univalent. The X bivalent segregates normally, and the Y univalent segregates randomly to one of the two poles. Which sperm genotypes could result from this specific meiotic process?
Answer: Only XY and X In this scenario, the XX bivalent segregates normally, so one X chromosome goes to each secondary spermatocyte. The Y univalent moves randomly to one of the poles. This results in two types of secondary spermatocytes: one containing an X and a Y chromosome, and one containing only an X chromosome. After meiosis II, the XY secondary spermatocyte produces two XY sperm. The X secondary spermatocyte produces two X sperm. Therefore, the resulting sperm genotypes are XY and X.
An organism has the genotype Aa Bb Cc. Gene A is on chromosome 1. Genes B and C are linked on chromosome 2, with the parental arrangement being BC/bc. The distance between B and C is 10 map units. What is the predicted frequency of gametes with the genotype a b C?
Answer: 2.5% This problem involves both independent assortment and linkage. First, determine the frequency of the 'a' allele. Since A is heterozygous (Aa), the probability of a gamete containing 'a' is 1/2. Second, determine the frequency of the 'b C' combination. B and C are linked, 10 m.u. apart, so the recombination frequency is 10%. The parental chromosomes are BC and bc. The gamete 'b C' is a recombinant type. The 10% recombination frequency is split between the two recombinant types (bC and Bc), so the frequency of 'b C' is 5% (or 0.05). Finally, since gene A assorts independently from B and C, multiply the probabilities: P(a b C) = P(a) P(b C) = 0.5 0.05 = 0.025, or 2.5%.
An allotetraploid plant was produced by crossing two different diploid species (Species 1, genome AA; Species 2, genome BB) and then doubling the chromosome number of the F1 hybrid. The resulting fertile allotetraploid has the genome AABB. Assuming normal bivalent pairing during meiosis, what is the chromosomal composition of the gametes it will produce?
Answer: AB When you encounter allopolyploidy questions, focus on how chromosomes pair during meiosis and what this means for gamete formation. Allotetraploids contain chromosome sets from two different species that maintain their distinct pairing preferences. In this cross, the initial diploid species have genomes AA and BB. Their F1 hybrid (AB) is typically sterile because the A and B chromosomes can't pair properly during meiosis—they're too different. However, when the chromosome number doubles to create the allotetraploid AABB, you now have homologous pairs: the A chromosomes can pair with each other, and the B chromosomes can pair with each other. During meiosis in the AABB allotetraploid, normal bivalent pairing occurs. The two A chromosomes form one bivalent, and the two B chromosomes form another bivalent. When these bivalents separate during meiosis I, each gamete receives one chromosome from each bivalent—meaning each gamete gets one A chromosome and one B chromosome, resulting in AB gametes. Choice A (A or B) represents what you'd expect from backcrossing to parental species, not from the allotetraploid itself. Choice B (AABB) would be the somatic cell composition, not a gamete—remember that meiosis reduces chromosome number by half. Choice C (AA or BB) incorrectly assumes that A and B chromosomes pair together and then separate as groups, which contradicts the principle of independent assortment. Remember: in allopolyploids, focus on how chromosome sets from different species pair independently, and always reduce the chromosome number by half for gametes.
In certain mouse populations, heterozygous males (+/t) for the t-haplotype exhibit meiotic drive, where approximately 90% of their functional sperm carry the t-allele. If such a male is crossed to a heterozygous female (+/t) who exhibits normal Mendelian segregation, what is the expected frequency of +/+ homozygotes among the viable progeny?
Answer: 5% This requires a modified Punnett square. The male produces sperm with frequencies: 0.90 t and 0.10 +. The female produces eggs with normal Mendelian frequencies: 0.50 t and 0.50 +. To find the frequency of +/+ progeny, we multiply the frequencies of the corresponding gametes: P(+/+ zygote) = P(+ sperm) × P(+ egg) = 0.10 × 0.50 = 0.05, or 5%.
An autotetraploid plant (4n) with genotype AAaa undergoes meiosis. During prophase I, the chromosomes form two bivalents. Assuming the only possible bivalent pairing is (A, a) and (a, a), and these bivalents segregate independently, what proportion of the resulting diploid (2n) gametes will have the genotype 'aa'?
Answer: 1/2 The plant forms two bivalents: (A,a) and (a,a). During anaphase I, each bivalent segregates its chromosomes to opposite poles. A resulting gamete will receive one chromosome from the (A,a) bivalent and one from the (a,a) bivalent. From the (a,a) bivalent, the gamete will always receive an 'a' allele (probability = 1). From the (A,a) bivalent, the gamete has a 1/2 chance of receiving 'A' and a 1/2 chance of receiving 'a'. To get a gamete with genotype 'aa', it must receive 'a' from the first bivalent AND 'a' from the second. The probability is 1/2 (from A,a pair) * 1 (from a,a pair) = 1/2.
An individual is a carrier for a balanced reciprocal translocation. Considering that adjacent-1 and alternate segregation from the quadrivalent intermediate occur at roughly equal frequencies (and adjacent-2 is rare), what proportion of this individual's viable gametes are expected to carry the translocation?
Answer: 50% In a translocation heterozygote, only gametes resulting from alternate segregation are genetically balanced and therefore viable. Adjacent-1 and adjacent-2 segregation patterns produce unbalanced, non-viable gametes. Alternate segregation produces two types of balanced gametes in equal proportions: one type containing a completely normal set of chromosomes, and another type containing the set of translocated chromosomes. Therefore, among the viable gametes, half will have the normal karyotype and half will carry the balanced translocation.
In a diploid organism, genes D and E are linked and 20 map units apart. An individual with the genotype De/dE undergoes meiosis. What is the expected frequency of gametes with the genotype DE?
Answer: 10% The parental genotype is De/dE, meaning the parental (non-recombinant) gametes are De and dE. The recombinant gametes are DE and de. A map distance of 20 m.u. corresponds to a 20% recombination frequency. This frequency is distributed equally between the two types of recombinant gametes. Therefore, the frequency of DE gametes is 20% / 2 = 10%, and the frequency of de gametes is also 10%.
A three-point test cross in Drosophila is set up to map three linked genes, x, y, and z. The genetic map is determined to be *x*—10 m.u.—*y*—20 m.u.—*z*. A separate experiment reveals that the female parent used in the cross is also heterozygous for a large paracentric inversion spanning the entire region between genes y and z. What is the expected, observable recombination frequency between y and z in the progeny of this cross?
Answer: 0% The presence of a heterozygous inversion acts as a crossover suppressor. While crossing over may physically occur within the inverted region, any chromatids that participate in a single crossover event within the inversion loop of a paracentric inversion will lead to the formation of dicentric and acentric products. Gametes receiving these broken chromosomes are non-viable and do not contribute to the progeny. As a result, no viable recombinant offspring for the genes within the inversion (y and z) will be recovered. The observable recombination frequency will therefore be 0% or very close to it.
An individual has a normal phenotype but is a carrier for a Robertsonian translocation involving chromosomes 13 and 14. This results in a karyotype with 45 chromosomes, including the fused t(13;14) chromosome. What is a possible chromosomal composition of a viable, genetically balanced gamete produced by this individual?
Answer: 22 chromosomes, including the t(13;14) chromosome. When you encounter Robertsonian translocation problems, focus on understanding how chromosome number changes and what "genetically balanced" means for gamete viability. A Robertsonian translocation fuses two acrocentric chromosomes at their centromeres, creating one large chromosome while losing the short arms. This individual has 45 chromosomes total: 43 normal chromosomes plus the fused t(13;14) chromosome. They're phenotypically normal because they still have all essential genetic material—just rearranged. During meiosis, this person can produce two types of genetically balanced gametes. The key insight is that "balanced" means having the correct amount of genetic material from chromosomes 13 and 14, regardless of how it's packaged. A gamete with 22 chromosomes including the t(13;14) chromosome contains all the essential genetic material from both chromosomes 13 and 14 fused together. This creates a viable, balanced gamete. Answer A (23 chromosomes with t(13;14)) would result in 46 total chromosomes after fertilization, but the offspring would have extra chromosome 13 and 14 material, causing imbalance. Answer B (22 chromosomes with normal 13 and 14) is impossible because this individual doesn't have separate normal copies of both chromosomes. Answer C (23 chromosomes with normal 13 and 14) has the same impossibility as B, plus would create trisomy. Remember: in Robertsonian translocations, balanced gametes can have either the fusion chromosome OR the separate normal chromosomes, but the total chromosome count must ensure genetic balance after fertilization. Focus on genetic content, not just chromosome number.
A human primary oocyte with genotype Aa undergoes nondisjunction of the homologous chromosomes during meiosis I. The resulting secondary oocyte receives both homologs. If this secondary oocyte proceeds through meiosis II, what are the expected genotypes of the mature ovum and the second polar body derived from it?
Answer: Ovum: Aa; Second polar body: Aa Meiosis I nondisjunction of homologs A and a results in a secondary oocyte that contains both, so its genotype is Aa (with each chromosome still duplicated). The first polar body is nullisomic. When this Aa secondary oocyte undergoes meiosis II, the sister chromatids separate. Due to unequal cytokinesis, the large ovum receives one chromatid of type A and one of type a, making its genotype Aa. The small second polar body also receives one chromatid of type A and one of type a, making its genotype Aa as well.
A woman is a carrier of a Robertsonian translocation, rob(14;21). She and her partner, who has a normal karyotype, wish to have a child. Considering only the three potential viable offspring outcomes (normal, balanced carrier, and trisomy 21), what is the theoretical risk of them having a child with translocation Down syndrome?
Answer: 1/3 A carrier of a rob(14;21) translocation can produce six types of gametes based on the segregation of chromosomes 14, 21, and the translocated chromosome. Three of these lead to non-viable monosomies (14 or 21) or trisomy 14 after fertilization. The three remaining gamete types lead to viable offspring: (1) a gamete with normal 14 and 21 chromosomes (leading to a phenotypically normal child), (2) a gamete with the rob(14;21) chromosome (leading to a balanced carrier like the mother), and (3) a gamete with the rob(14;21) chromosome plus a normal chromosome 21 (leading to translocation Down syndrome). Assuming these three viable outcomes are equally probable, the risk is 1/3.

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