Genetics Quiz: Meiosis And Genetic Variation

What this quiz covers

This quiz focuses on Meiosis And Genetic Variation, 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. 20 pg and 20 pg (correct answer)
2. 20 pg and 10 pg
3. 10 pg and 10 pg
4. 40 pg and 20 pg

Explanation: A G1 somatic cell is diploid (2n) and has a 2C amount of DNA. So, 2C = 20 pg. Before meiosis, DNA replication occurs, so a primary oocyte entering meiosis I has a 4C amount of DNA (40 pg). Meiosis I separates homologous chromosomes, halving both the chromosome number (to n) and the DNA content (to 2C). Although cytokinesis is unequal, the nuclear division is equal. Therefore, both the secondary oocyte and the first polar body are haploid (n) but contain chromosomes consisting of two sister chromatids, resulting in a 2C DNA content for each, which is 20 pg.

Question 2

1. Meiosis would proceed normally, but all resulting gametes would be diploid.
2. The cell would arrest in prophase I and be unable to form bivalents.
3. Homologous chromosomes would segregate normally in meiosis I, but sister chromatids would segregate randomly in meiosis II.
4. Massive aneuploidy in all resulting gametes due to premature separation of sister chromatids and random segregation during meiosis I. (correct answer)

Explanation: Sister chromatid cohesion is essential for maintaining the structure of the bivalent after crossing over. Cohesion distal to the chiasma holds the homologous chromosomes together. If cohesion were lost before meiosis I, the sister chromatids would separate. The four chromatids of a homologous pair would no longer form a stable bivalent. This would lead to improper attachment of microtubules and random segregation of the four chromatids during meiosis I, resulting in gametes with severe aneuploidy.

Question 3

1. Two gametes with one large, one medium, and one small; two gametes with one large and one small.
2. Two gametes with one large, two medium, and one small; two gametes with one large and one small. (correct answer)
3. Four gametes, each with one large, one medium, and one small chromosome.
4. One gamete with two large, two medium, and two small; three gametes with no chromosomes.

Explanation: In meiosis I non-disjunction, the medium-sized homologous pair fails to separate. One secondary gametocyte receives both medium chromosomes, plus one large and one small (L, M, M, S). The other secondary gametocyte receives only the large and small chromosomes (L, S). Both cells then proceed to meiosis II, where sister chromatids separate. The first cell (L, M, M, S) produces two gametes, each with one large, two medium, and one small chromosome. The second cell (L, S) produces two gametes, each with one large and one small chromosome. This results in two n+1 gametes and two n-1 gametes.

Question 4

1. Two gametes of genotype AaBb and two gametes lacking the chromosome.
2. Two gametes of genotype AB, two of genotype ab, and two lacking the chromosome.
3. Two gametes of genotype AABB, two of genotype aabb, and two lacking the chromosome.
4. Two gametes of genotype AB/ab and two gametes lacking the chromosome. (correct answer)

Explanation: Non-disjunction in Meiosis I involves the failure of homologous chromosomes to separate. The individual has one chromosome with alleles A and B, and its homolog with alleles a and b. In this event, one secondary gametocyte receives both homologous chromosomes (constitution AB/ab), while the other receives none. The cell with the extra chromosomes then proceeds to Meiosis II, where sister chromatids separate. This is effectively a mitotic division, producing two diploid gametes, each containing one AB chromosome and one ab chromosome (genotype AB/ab). The other secondary gametocyte produces two nullisomic gametes lacking this chromosome.

Question 5

1. Meiosis would proceed normally, but all resulting gametes would be diploid.
2. The cell would arrest in prophase I and be unable to form bivalents.
3. Homologous chromosomes would segregate normally in meiosis I, but sister chromatids would segregate randomly in meiosis II.
4. Massive aneuploidy in all resulting gametes due to premature separation of sister chromatids and random segregation during meiosis I. (correct answer)

Explanation: Sister chromatid cohesion is essential for maintaining the structure of the bivalent after crossing over. Cohesion distal to the chiasma holds the homologous chromosomes together. If cohesion were lost before meiosis I, the sister chromatids would separate. The four chromatids of a homologous pair would no longer form a stable bivalent. This would lead to improper attachment of microtubules and random segregation of the four chromatids during meiosis I, resulting in gametes with severe aneuploidy.

Question 6

1. The species must reproduce asexually, as sexual reproduction is impossible without crossing over.
2. All gametes produced by an individual will be genetically identical to each other.
3. Genetic variation among gametes will be generated only by independent assortment of homologous chromosomes. (correct answer)
4. Mendel's Law of Segregation would be violated, but the Law of Independent Assortment would be unaffected.

Explanation: Meiosis has two main mechanisms for generating genetic variation: crossing over and independent assortment. If crossing over is absent, the sister chromatids of each chromosome remain identical. However, as long as the species has more than one pair of chromosomes, the independent assortment of non-homologous chromosomes during metaphase I will still shuffle the parental chromosomes, creating genetically diverse gametes. For example, in an organism with genotype AaBb on different chromosomes, gametes AB, Ab, aB, and ab can still be formed. Therefore, genetic variation is reduced but not eliminated. Sexual reproduction is still possible, and Mendel's laws would not be violated.

Question 7

1. A significant reduction in chiasma formation and an increase in chromosome non-disjunction. (correct answer)
2. An increase in the frequency of crossing over, leading to greater genetic diversity.
3. Meiosis will proceed normally, as crossing over can occur without the synaptonemal complex.
4. The cell will arrest in G2 phase and will not be able to enter meiosis.

Explanation: The synaptonemal complex acts as a scaffold that aligns homologous chromosomes precisely, which is essential for the proper repair of double-strand breaks as crossovers. Without this complex, homologs may not align correctly, and the repair process is much less likely to result in a crossover that forms a chiasma. Chiasmata are crucial for physically linking homologous chromosomes, ensuring they orient correctly on the metaphase I plate. A lack of chiasmata leads to a high frequency of univalents, which segregate randomly, causing widespread non-disjunction and aneuploidy.

Question 8

1. The efficiency of DNA replication during the S phase prior to meiosis.
2. The formation of the synaptonemal complex during prophase I.
3. The maintenance of sister chromatid cohesion during the prolonged prophase I arrest. (correct answer)
4. The breakdown of the nuclear envelope at the end of prophase I.

Explanation: Human oocytes are arrested in prophase I from fetal development until ovulation, a period that can last for decades. During this long arrest, the cohesin proteins that hold sister chromatids together can degrade. Weakened cohesion can lead to premature separation of sister chromatids or unstable bivalents, both of which increase the likelihood of incorrect chromosome segregation (non-disjunction) during meiosis I, which is the leading cause of age-related trisomies.

Question 9

1. Prophase I
2. Metaphase I
3. Prophase II
4. Metaphase II (correct answer)

Explanation: Human oogenesis involves two meiotic arrests. The first occurs in Prophase I, where primary oocytes remain from fetal development until puberty. After completing Meiosis I, the resulting secondary oocyte begins Meiosis II but arrests again, this time in Metaphase II. It is in this Metaphase II-arrested state that the oocyte is ovulated. Fertilization provides the signal for the oocyte to complete Meiosis II, extruding the second polar body and forming the mature ovum.

Question 10

1. 32 distinct gametes; a probability of 1/32. (correct answer)
2. 1024 distinct gametes; a probability of 1/1024.
3. 32 distinct gametes; a probability of 1/10.
4. 10 distinct gametes; a probability of 1/32.

Explanation: The number of genetically distinct gametes produced by independent assortment is given by the formula $2^n$, where n is the haploid number of chromosomes. If $2n=10$, then $n=5$. Therefore, the number of distinct gametes is $2^5 = 32$. For any given homologous pair, the probability that a gamete will receive the paternal chromosome is 1/2. Since the assortment of the 5 pairs is independent, the probability of a gamete receiving only paternal chromosomes for all 5 pairs is $(1/2)^5 = 1/32$.

Question 11

1. The law of segregation applied to each gene locus individually.
2. The law of independent assortment regarding the alignment of homologous chromosomes. (correct answer)
3. Crossing over between the G and H gene loci during prophase I.
4. The separation of sister chromatids during meiosis II.

Explanation: The production of four different gamete types from a dihybrid demonstrates Mendel's Law of Independent Assortment. Since the genes are on different chromosomes, the homologous pair carrying G/g aligns at the metaphase I plate independently of the homologous pair carrying H/h. This results in two equally likely alignments: one that produces GH and gh gametes, and one that produces Gh and gH gametes. Crossing over (C) is irrelevant for genes on different chromosomes. Segregation (A) explains why each gamete gets G or g (and H or h), but not how they are combined. Separation of sister chromatids (D) does not create new combinations of alleles from different genes.

Question 12

1. 7.5 map units
2. 15 map units (correct answer)
3. 30 map units
4. 70 map units

Explanation: The 4:4 pattern represents first-division segregation (FDS), where no crossover occurred between the gene and the centromere. The 2:2:2:2 and 2:4:2 patterns represent second-division segregation (SDS), indicating a crossover occurred. The frequency of SDS asci is 30/100 = 30%. The map distance from the centromere is calculated as $(1/2 \times \%SDS)$. The 1/2 factor is included because in any meiotic event with a single crossover, only two of the four chromatids are recombinant. Therefore, the map distance is $1/2 \times 30\% = 15\%$, which corresponds to 15 map units.

Question 13

1. Independent assortment of genes Y and R on non-homologous chromosomes.
2. Complete linkage of genes Y and R with no meiotic crossing over.
3. Linkage of genes Y and R, with crossing over occurring in a fraction of meioses. (correct answer)
4. A lethal allele interaction that is skewing the expected Mendelian ratios.

Explanation: In this test cross, the parental phenotypes (derived from the original cross to make the F1, likely YYRR x yyrr) are Y_R_ and yyrr. The recombinant phenotypes are Y_rr and yyR_. The number of parental progeny is 435 + 425 = 860. The number of recombinant progeny is 72 + 68 = 140. The total is 1000. The recombination frequency is 140/1000 = 14%. Since this frequency is significantly less than the 50% expected for independent assortment (which would yield a 1:1:1:1 ratio), the genes are linked. The presence of recombinants indicates that crossing over occurs between them.

Question 14

1. Independent assortment of genes Y and R on non-homologous chromosomes.
2. Complete linkage of genes Y and R with no meiotic crossing over.
3. Linkage of genes Y and R, with crossing over occurring in a fraction of meioses. (correct answer)
4. A lethal allele interaction that is skewing the expected Mendelian ratios.

Explanation: In this test cross, the parental phenotypes (derived from the original cross to make the F1, likely YYRR x yyrr) are Y_R_ and yyrr. The recombinant phenotypes are Y_rr and yyR_. The number of parental progeny is 435 + 425 = 860. The number of recombinant progeny is 72 + 68 = 140. The total is 1000. The recombination frequency is 140/1000 = 14%. Since this frequency is significantly less than the 50% expected for independent assortment (which would yield a 1:1:1:1 ratio), the genes are linked. The presence of recombinants indicates that crossing over occurs between them.

Question 15

1. 1 AA : 2 Aa : 1 aa
2. 3 A : 1 a
3. 1 AA : 4 Aa : 1 aa (correct answer)
4. 1 AA : 1 aa

Explanation: A tetraploid (4n) organism produces diploid (2n) gametes. The parent genotype is AAaa. During meiosis, the four homologous chromosomes will segregate, with two moving to each pole to form the diploid gamete. We can determine the combinations by considering picking two chromosomes from the set {A, A, a, a}. There are $\binom{4}{2} = 6$ possible combinations of two chromosomes. These are: one {A, A} combination, one {a, a} combination, and four distinct {A, a} combinations. Therefore, the ratio of gamete genotypes will be 1 AA : 4 Aa : 1 aa.

Question 16

1. The separation of homologous chromosomes during anaphase I.
2. The separation of sister chromatids during anaphase II.
3. The random alignment of homologous pairs at the metaphase plate during metaphase I. (correct answer)
4. The crossing over between non-sister chromatids during prophase I.

Explanation: The Law of Independent Assortment states that alleles for different genes assort independently of one another during gamete formation. This occurs because the orientation of each homologous pair (bivalent) at the metaphase I plate is random and independent of the orientation of all other pairs. This random alignment determines which pole the maternal and paternal chromosomes will travel to, leading to different combinations of chromosomes in the resulting gametes. The separation in anaphase I (A) is the basis for the Law of Segregation. Separation in anaphase II (B) splits identical chromatids (barring crossover). Crossing over (D) creates new allele combinations on a single chromosome but is not the primary mechanism for the assortment of entire chromosomes.

Question 17

1. 32 distinct gametes; a probability of 1/32. (correct answer)
2. 1024 distinct gametes; a probability of 1/1024.
3. 32 distinct gametes; a probability of 1/10.
4. 10 distinct gametes; a probability of 1/32.

Explanation: The number of genetically distinct gametes produced by independent assortment is given by the formula $2^n$, where n is the haploid number of chromosomes. If $2n=10$, then $n=5$. Therefore, the number of distinct gametes is $2^5 = 32$. For any given homologous pair, the probability that a gamete will receive the paternal chromosome is 1/2. Since the assortment of the 5 pairs is independent, the probability of a gamete receiving only paternal chromosomes for all 5 pairs is $(1/2)^5 = 1/32$.

Question 18

1. Two gametes of genotype AaBb and two gametes lacking the chromosome.
2. Two gametes of genotype AB, two of genotype ab, and two lacking the chromosome.
3. Two gametes of genotype AABB, two of genotype aabb, and two lacking the chromosome.
4. Two gametes of genotype AB/ab and two gametes lacking the chromosome. (correct answer)

Explanation: Non-disjunction in Meiosis I involves the failure of homologous chromosomes to separate. The individual has one chromosome with alleles A and B, and its homolog with alleles a and b. In this event, one secondary gametocyte receives both homologous chromosomes (constitution AB/ab), while the other receives none. The cell with the extra chromosomes then proceeds to Meiosis II, where sister chromatids separate. This is effectively a mitotic division, producing two diploid gametes, each containing one AB chromosome and one ab chromosome (genotype AB/ab). The other secondary gametocyte produces two nullisomic gametes lacking this chromosome.

Question 19

1. 1 AA : 2 Aa : 1 aa
2. 3 A : 1 a
3. 1 AA : 4 Aa : 1 aa (correct answer)
4. 1 AA : 1 aa

Explanation: A tetraploid (4n) organism produces diploid (2n) gametes. The parent genotype is AAaa. During meiosis, the four homologous chromosomes will segregate, with two moving to each pole to form the diploid gamete. We can determine the combinations by considering picking two chromosomes from the set {A, A, a, a}. There are $\binom{4}{2} = 6$ possible combinations of two chromosomes. These are: one {A, A} combination, one {a, a} combination, and four distinct {A, a} combinations. Therefore, the ratio of gamete genotypes will be 1 AA : 4 Aa : 1 aa.

Question 20

1. The law of segregation applied to each gene locus individually.
2. The law of independent assortment regarding the alignment of homologous chromosomes. (correct answer)
3. Crossing over between the G and H gene loci during prophase I.
4. The separation of sister chromatids during meiosis II.

Explanation: The production of four different gamete types from a dihybrid demonstrates Mendel's Law of Independent Assortment. Since the genes are on different chromosomes, the homologous pair carrying G/g aligns at the metaphase I plate independently of the homologous pair carrying H/h. This results in two equally likely alignments: one that produces GH and gh gametes, and one that produces Gh and gH gametes. Crossing over (C) is irrelevant for genes on different chromosomes. Segregation (A) explains why each gamete gets G or g (and H or h), but not how they are combined. Separation of sister chromatids (D) does not create new combinations of alleles from different genes.