- An organism has a diploid chromosome number of 10 (2n=10). Assume no crossing over occurs. How many genetically distinct gametes can this organism produce solely through independent assortment, and what is the probability of producing a gamete that contains only chromosomes of paternal origin?
Answer: 32 distinct gametes; a probability of 1/32.
The number of genetically distinct gametes produced by independent assortment is given by the formula 2n, where n is the haploid number of chromosomes. If 2n=10, then n=5. Therefore, the number of distinct gametes is 25=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. - In a certain strain of Drosophila, a large chromosomal inversion suppresses crossing over across the entire length of chromosome 2 in heterozygous females. An AABB male is crossed with an aabb female from this strain, where genes A and B are both on chromosome 2. An F1 female is then test-crossed with an aabb male. What phenotypic ratio is expected in the F2 progeny?
Answer: 1 dominant A, dominant B : 1 recessive a, recessive b
The F1 female has the genotype AaBb, with the AB alleles on one chromosome and the ab alleles on the homologous chromosome. The chromosomal inversion suppresses crossing over, meaning these alleles cannot be recombined. The F1 female can therefore only produce two types of gametes: parental gametes containing AB and parental gametes containing ab. When test-crossed with an aabb male (who only produces ab gametes), the resulting F2 progeny will be AaBb (dominant A, dominant B phenotype) and aabb (recessive a, recessive b phenotype) in a 1:1 ratio. - In meiosis, the synaptonemal complex is responsible for holding homologous chromosomes together during prophase I to facilitate crossing over. Suppose a mutation prevents the formation of the synaptonemal complex, but does not affect the cell's ability to create double-strand breaks. What is the most likely outcome?
Answer: A significant reduction in chiasma formation and an increase in chromosome non-disjunction.
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.
- In a diploid organism, genes A and B are linked on the same chromosome. An individual with genotype AB/ab undergoes meiosis. If non-disjunction of this chromosome pair occurs during Meiosis I, what will be the allelic constitution of the resulting aneuploid gametes?
Answer: Two gametes of genotype AB/ab and two gametes lacking the chromosome.
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.
- In a beetle, recessive alleles a, b, and c are on the same chromosome. A test cross of a trihybrid female (heterozygous for all three genes) with a homozygous recessive male produces 1000 offspring with the phenotypes shown in the table. Based on these data, what is the most likely reason for the absence of the a B c and A b C phenotypic classes?
Answer: The female's parental chromosomes were A b c and a B C.
The most frequent phenotypic classes in the offspring of a test cross represent the parental gametes from the heterozygous parent. Here, the most frequent classes are A b c (405) and a B C (415). This indicates that the female's original, non-recombinant chromosomes carried these allele combinations. Double crossover events would produce the least frequent classes, which are A B C and a b c. The absence of the a B c and A b C classes (phenotypes are lowercase for recessive) means those were the expected parental types if the allele arrangement was different, which is incorrect.
- Which of the following statements represents the most fundamental distinction between the chromosomal behavior in anaphase I and anaphase II of meiosis?
Answer: Homologous chromosomes separate in anaphase I, while sister chromatids separate in anaphase II.
The defining event of anaphase I is the separation of homologous chromosomes, which are pulled to opposite poles. Each chromosome still consists of two sister chromatids. This is the reductional division. In contrast, the defining event of anaphase II is the separation of sister chromatids, which are then considered individual chromosomes. This is an equational division, analogous to mitotic anaphase.
- A cell from a species with three pairs of chromosomes (2n=6) enters meiosis. The chromosomes are distinguished by size: large, medium, and small. If non-disjunction of the medium-sized homologous chromosomes occurs during meiosis I, what is the chromosome content of the four gametes produced at the end of meiosis II?
Answer: Two gametes with one large, two medium, and one small; two gametes with one large and one small.
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.
- In a fungus that produces ordered tetrads, a cross is made between a strain requiring arginine (arg−) and a wild-type strain (arg+). Analysis of 100 asci reveals 70 with a 4:4 allele pattern and 30 with a 2:2:2:2 or 2:4:2 pattern. What is the approximate map distance between the arg gene and the centromere?
Answer: 15 map units
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×%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 1/2×30%=15%, which corresponds to 15 map units. - A diploid organism has a genome size of 20 picograms (pg) in a G1 somatic cell. What is the expected DNA content of a secondary oocyte and the first polar body, respectively, immediately after the completion of meiosis I?
Answer: 20 pg and 20 pg
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.
- A geneticist is studying two genes, Y and R, in a plant. A test cross of a dihybrid F1 individual (YyRr) yields the following F2 progeny: 435 plants with Y and R dominant phenotypes, 72 with Y dominant and r recessive, 68 with y recessive and R dominant, and 425 with y and r recessive phenotypes. Which meiotic phenomenon best explains this result?
Answer: Linkage of genes Y and R, with crossing over occurring in a fraction of meioses.
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.
- Imagine a hypothetical mutation that causes sister chromatid cohesion to be completely lost before meiosis I begins, but leaves homologous chromosome pairing and synapsis intact. What would be the most probable outcome of meiosis in an individual with this mutation?
Answer: Massive aneuploidy in all resulting gametes due to premature separation of sister chromatids and random segregation during meiosis I.
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.
- Which of the following events is the primary source of the extensive genetic variation observed in the gametes produced by a single individual?
Answer: The shuffling of parental alleles through independent assortment and crossing over.
While all options relate to genetics, the question asks for the source of variation in the gametes of a single individual. Random fertilization (B) creates variation in offspring by combining gametes from two individuals. Somatic mutations (C) are not heritable. Allele segregation in meiosis II (A) is a consequence of processes in meiosis I. The primary source of variation among the gametes from one parent is the combination of independent assortment (random orientation of homologous pairs in metaphase I) and crossing over (exchange of segments between homologous chromosomes in prophase I). These two meiotic mechanisms create new combinations of the individual's parental alleles.
- The incidence of trisomy 21 (Down syndrome) increases significantly with maternal age. This is primarily attributed to age-related degradation of factors involved in meiosis. Which specific meiotic process is thought to be most affected by this degradation, leading to an increased risk of non-disjunction?
Answer: The maintenance of sister chromatid cohesion during the prolonged prophase I arrest.
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.
- If a diploid cell with genotype GgHh undergoes meiosis, and the genes are on different chromosomes, the resulting gametes can be GH, Gh, gH, or gh. The formation of gametes with genotypes Gh and gH is a direct result of:
Answer: The law of independent assortment regarding the alignment of homologous chromosomes.
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.
- A stable autotetraploid plant (4n) has the genotype AAaa for a single gene. Assuming random chromosome segregation (any two of the four homologous chromosomes can migrate to a pole), what is the expected ratio of genotypes in the diploid (2n) gametes produced by this plant?
Answer: 1 AA : 4 Aa : 1 aa
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 (24​)=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. - A botanist discovers a new plant species where meiosis consistently proceeds without any crossing over. Which of the following statements accurately describes the genetic consequences for this species?
Answer: Genetic variation among gametes will be generated only by independent assortment of homologous chromosomes.
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.
- During oogenesis in humans, meiosis is arrested twice at specific stages. A secondary oocyte is ovulated and, if fertilized, completes meiosis II. At the moment of fertilization, the secondary oocyte is arrested in which stage?
Answer: Metaphase II
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.
- Mendel's Law of Independent Assortment is a direct consequence of which of the following meiotic events?
Answer: The random alignment of homologous pairs at the metaphase plate during metaphase I.
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.
