Genetics Quiz: Gamete Outcomes From Meiosis

What this quiz covers

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

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

Question 1

1. Sperm with genotypes AB/ab and nullisomic sperm.
2. Sperm with genotypes AB, ab, Ab, and aB.
3. Sperm with genotypes ab, AB/AB, and nullisomic for the chromosome. (correct answer)
4. Sperm with genotypes AB and ab only.

Explanation: 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.

Question 2

1. 25%
2. 50% (correct answer)
3. 75%
4. 100%

Explanation: 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.

Question 3

1. 5%
2. 10% (correct answer)
3. 20%
4. 40%

Explanation: 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%.

Question 4

1. 1/6
2. 1/4
3. 1/3
4. 1/2 (correct answer)

Explanation: 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.

Question 5

1. 23 chromosomes, including the t(13;14) chromosome.
2. 22 chromosomes, including a normal 13 and a normal 14.
3. 23 chromosomes, including a normal 13 and a normal 14.
4. 22 chromosomes, including the t(13;14) chromosome. (correct answer)

Explanation: 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.

Question 6

1. Ovum: X^AX^A; First polar body: X^a; Second polar body: nullisomic (correct answer)
2. Ovum: X^AX^a; First polar body: nullisomic; Second polar body: X^AX^a
3. Ovum: X^A; First polar body: X^a; Second polar body: X^a
4. Ovum: nullisomic; First polar body: X^AX^a; Second polar body: nullisomic

Explanation: 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.

Question 7

1. Only X and Y
2. Only XY and X (correct answer)
3. Only XX and Y
4. Only XX and YY

Explanation: 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.

Question 8

1. A second division segregation pattern, resulting from a crossover between the arg gene and the centromere.
2. A first division segregation pattern, indicating no crossover occurred between the arg gene and the centromere. (correct answer)
3. Gene conversion at the arg locus, resulting in a non-Mendelian 6:2 ratio of spores.
4. Nondisjunction during meiosis I, leading to an abnormal number of chromosomes in the spores.

Explanation: 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.

Question 9

1. A or B
2. AABB
3. AA or BB
4. AB (correct answer)

Explanation: 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.

Question 10

1. Two non-recombinant gametes (one normal, one inverted) and two recombinant gametes containing dicentric and acentric chromosomes.
2. Two non-recombinant gametes (one normal, one inverted) and two non-viable recombinant gametes, each with a duplication and a deletion. (correct answer)
3. Four viable gametes: two non-recombinant parental types and two recombinant types with new allele combinations.
4. Four non-viable gametes due to the universal formation of chromosomes with duplications and deletions.

Explanation: 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.

Question 11

1. 5% (correct answer)
2. 10%
3. 25%
4. 50%

Explanation: 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%.

Question 12

1. 0% (correct answer)
2. 10%
3. 20%
4. 30%

Explanation: 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.

Question 13

1. 1.25%
2. 2.5% (correct answer)
3. 5.0%
4. 22.5%

Explanation: 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%.

Question 14

1. Ovum: AA; Second polar body: aa
2. Ovum: A; Second polar body: a
3. Ovum: Aa; Second polar body: Aa (correct answer)
4. Ovum: Aa; Second polar body: nullisomic

Explanation: 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.

Question 15

1. Black body, vestigial wings
2. Gray body, vestigial wings
3. Black body, normal wings
4. Gray body, normal wings (correct answer)

Explanation: 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.

Question 16

1. Ovum: X^AX^A; First polar body: X^a; Second polar body: nullisomic (correct answer)
2. Ovum: X^AX^a; First polar body: nullisomic; Second polar body: X^AX^a
3. Ovum: X^A; First polar body: X^a; Second polar body: X^a
4. Ovum: nullisomic; First polar body: X^AX^a; Second polar body: nullisomic

Explanation: 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.

Question 17

1. 25%
2. 50% (correct answer)
3. 75%
4. 100%

Explanation: 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.

Question 18

1. 1/2
2. 1/3 (correct answer)
3. 1/4
4. 1/6

Explanation: 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.

Question 19

1. 1/6
2. 1/4
3. 1/3
4. 1/2 (correct answer)

Explanation: 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.

Question 20

1. 5% (correct answer)
2. 10%
3. 25%
4. 50%

Explanation: 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%.