Meiosis — AP Biology Study Guide

For: AP Biology candidates sitting AP Biology.

Covers: Meiosis definition, stages of meiosis I and II, ploidy changes, crossing over, independent assortment, comparison of meiosis to mitosis, sources of genetic variation, and nondisjunction errors, aligned to AP Biology CED Unit 5.

You should already know: Basics of the eukaryotic cell cycle, stages of mitosis, definition of ploidy.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Biology style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Meiosis?

Meiosis is a type of eukaryotic cell division that produces genetically distinct haploid gametes (or spores, in plants and fungi) from a diploid parental germ cell. It is the required foundation for all sexual reproduction, and accounts for ~10-15% of Unit 5 Heredity exam weight, which translates to ~8-11% of the total AP Biology exam score. Meiosis is tested across both multiple-choice (MCQ) and free-response (FRQ) sections: it commonly appears as a standalone MCQ, or as the conceptual base for FRQs that connect to Mendelian genetics, non-Mendelian inheritance, or evolutionary variation.

Standard AP notation uses $n$ to represent the haploid number (number of distinct chromosomes in one set), and $2n$ for diploid (two sets, one inherited from each parent). A common synonym for meiosis seen in exam questions is "reduction division", referring to the reduction in ploidy that occurs during the first division. Unlike mitosis, which produces genetically identical diploid somatic cells for growth and repair, meiosis generates genetic variation that acts as the raw material for natural selection. It also maintains a constant ploidy across generations: fertilization combines two haploid gametes to restore diploidy in the zygote, so meiosis must halve ploidy before gamete formation.

2. Stages of Meiosis and Ploidy Changes

Meiosis consists of two consecutive cell divisions (meiosis I and meiosis II) that follow a single round of DNA replication during interphase. The defining ploidy change occurs in meiosis I, which is why it is called the reduction division. After interphase, all chromosomes are replicated: each chromosome has two identical sister chromatids connected at the centromere, but ploidy remains $2n$, because the number of distinct chromosome sets has not changed—only the total DNA content per chromosome has doubled.

In meiosis I: Prophase I sees homologous chromosomes pair up (a process called synapsis) and crossing over occurs between non-sister chromatids. In metaphase I, homologous chromosome pairs line up independently at the metaphase plate. In anaphase I, homologous chromosomes separate, while sister chromatids remain attached at the centromere. After cytokinesis I, two haploid cells form: each cell has only one set of chromosomes (no homologous pairs remaining), so ploidy drops from $2n$ to $n$, even though each chromosome still has two sister chromatids. Meiosis II is nearly identical to mitosis: it separates sister chromatids, resulting in four genetically distinct haploid daughter cells, with no further change in ploidy.

Worked Example

A diploid somatic cell from a coastal redwood has $2n=16$. What is the number of chromosomes, number of chromatids, and ploidy of a cell at the end of prophase I, and of a cell at the end of meiosis II?

1. Start with baseline values: $2n=16$ means there are 8 distinct chromosomes per haploid set. After interphase, all chromosomes are replicated, so ploidy remains $2n$.
2. Prophase I occurs before the first meiotic division, so all homologous chromosomes are still in the same cell. Number of chromosomes (counted by centromeres) = 16, number of chromatids = $16\times 2=32$, ploidy = $2n$.
3. Meiosis I reduces ploidy by half, producing two cells with 8 replicated chromosomes each (each with two chromatids). Meiosis II separates sister chromatids, with no change in ploidy.
4. After meiosis II and cytokinesis, each daughter cell has 8 unreplicated chromosomes (one chromatid each), for 8 total chromosomes, and is haploid ($n$).
5. Final result: End of prophase I = 16 chromosomes, 32 chromatids, $2n$; End of meiosis II = 8 chromosomes, 8 chromatids, $n$.

Exam tip: On AP MCQs, always check if the question asks for number of chromosomes or number of DNA molecules (chromatids). Chromosome count is always based on number of centromeres, so a replicated chromosome with two sister chromatids counts as one chromosome, not two.

3. Sources of Genetic Variation from Meiosis

Three key events during and after meiosis generate genetic variation in sexually reproducing populations, which is the foundation of heredity and evolution. First, crossing over (homologous recombination) during prophase I exchanges homologous segments of DNA between non-sister chromatids of homologous chromosomes. This creates new combinations of alleles on each chromosome that were not inherited from either parent, making every chromosome produced by meiosis genetically unique.

Second, independent assortment of homologous chromosomes during metaphase I: each pair of homologous chromosomes lines up at the metaphase plate independently of every other pair, meaning which pole a maternal or paternal chromosome moves to does not affect any other pair. The number of distinct combinations of maternal and paternal chromosomes possible in gametes (assuming no crossing over) is given by the formula: $$\text{Number of gamete combinations}=2^n$$ where $n$ is the haploid number of the organism. Third, random fertilization adds even more variation: any sperm can fuse with any egg, so the number of combinations in the zygote becomes $2^n\times 2^n=4^n$.

Worked Example

A diploid insect species has $2n=12$. How many distinct combinations of maternal and paternal chromosomes are possible in the species' gametes, assuming no crossing over occurs?

1. Confirm the question asks for combinations from independent assortment alone, with no crossing over, so we use the $2^n$ formula.
2. Calculate the haploid number $n$ from the given diploid number: $2n=12⟹n=6$.
3. Substitute into the formula: $2^6=64$.
4. If crossing over were included, the total number of genetically distinct gametes would be far higher, since recombination creates new allele combinations on individual chromosomes. Final result: 64 distinct combinations of maternal and paternal chromosomes are possible.

Exam tip: When the question asks for the number of genetic combinations in a zygote (after fertilization), remember to multiply the combinations from two independent gametes, so the formula becomes $4^n$, not $2^n$.

4. Meiosis Comparison to Mitosis and Nondisjunction Errors

Meiosis and mitosis share core spindle-based cell division machinery, but have key functional differences that are frequently tested on the AP exam. Mitosis occurs in somatic (body) cells, produces two genetically identical diploid daughter cells, has one round of division, and no synapsis or crossing of homologous chromosomes. Meiosis occurs exclusively in germ cells (cells that produce gametes), produces four genetically distinct haploid daughter cells, has two rounds of division, and has synapsis and crossing over in prophase I.

The most common error in meiosis is nondisjunction: the failure of chromosomes to separate properly during anaphase. Nondisjunction can occur in anaphase I (failure of homologous chromosomes to separate) or anaphase II (failure of sister chromatids to separate). Nondisjunction produces aneuploid gametes (gametes with an abnormal number of chromosomes): one gamete gets an extra copy of the affected chromosome ($n+1$), and the other gets no copy ($n-1$). When an aneuploid gamete fuses with a normal gamete during fertilization, the resulting zygote is either trisomic ($2n+1$, three copies of the chromosome) or monosomic ($2n-1$, one copy). Most aneuploidies are lethal in early development, but some are viable, like trisomy 21 in humans.

Worked Example

Nondisjunction of chromosome 9 occurs during anaphase I of meiosis in a diploid sunflower that has $2n=34$. What are the chromosome numbers of the four resulting pollen grains (male gametes) produced from this parent cell?

1. Normal meiosis I produces two daughter cells with 17 chromosomes each, which then divide to produce four 17-chromosome pollen grains.
2. When nondisjunction occurs in anaphase I, both homologous copies of chromosome 9 move to the same pole of the cell. This produces one daughter cell with 18 chromosomes (extra chromosome 9) and one daughter cell with 16 chromosomes (missing chromosome 9).
3. Meiosis II separates sister chromatids normally, so both daughter cells from the first abnormal cell retain their chromosome number, and both from the second abnormal cell do as well.
4. Final result: two pollen grains have 18 chromosomes, two pollen grains have 16 chromosomes; no pollen grains have the normal 17 chromosomes.

Exam tip: If nondisjunction occurs in anaphase II, only two of the four gametes will be abnormal (one $n+1$, one $n-1$) and two will be normal. Always confirm whether the question places nondisjunction in meiosis I or meiosis II.

5. Common Pitfalls (and how to avoid them)

• Wrong move: Counting a replicated chromosome with two sister chromatids as two separate chromosomes. Why: Students confuse number of DNA molecules (chromatids) with number of chromosomes, which is defined by number of centromeres. Correct move: Always count centromeres to get chromosome number, regardless of whether the chromosome is replicated.
• Wrong move: Claiming sister chromatids separate during anaphase I of meiosis. Why: Students mix up the separation events of meiosis I and meiosis II, and confuse meiosis with mitosis. Correct move: Memorize the rule: Meiosis I = homologous chromosomes separate; Meiosis II = sister chromatids separate.
• Wrong move: Stating that ploidy is halved during meiosis II. Why: Students see two cell divisions and incorrectly assume ploidy is halved both times. Correct move: Ploidy reduction only occurs in meiosis I; meiosis II only separates sister chromatids, so ploidy does not change.
• Wrong move: Claiming crossing over occurs between sister chromatids of the same chromosome. Why: Students mix up the identity of chromatids in homologous pairs. Correct move: Crossing over always occurs between non-sister chromatids of homologous chromosomes, to create new allele combinations.
• Wrong move: Stating that only two gametes are abnormal when nondisjunction occurs in meiosis I. Why: Students generalize the outcome of meiosis II nondisjunction to meiosis I. Correct move: Nondisjunction in meiosis I = all four gametes abnormal; nondisjunction in meiosis II = two gametes abnormal.
• Wrong move: Using $2^n$ to calculate the number of genetic combinations in a zygote. Why: Students forget that fertilization combines two independent gametes. Correct move: Gamete combinations = $2^n$; zygote combinations = $2^n\times 2^n=4^n$; always confirm what the question is asking for.

6. Practice Questions (AP Biology Style)

Question 1 (Multiple Choice)

A diploid moss species has a haploid number of 12. How many chromosomes would be present in a single moss spore produced by meiosis, and what is the ploidy of the spore? A) 12 chromosomes, diploid (2n) B) 24 chromosomes, haploid (n) C) 12 chromosomes, haploid (n) D) 24 chromosomes, diploid (2n)

Worked Solution: The question states the haploid number is 12, so diploid somatic cells are $2n=24$. Spores are reproductive cells produced by meiosis from diploid sporophyte moss cells, so meiosis produces haploid spores. A haploid cell has $n$ chromosomes, which equals 12 here. This eliminates options A (wrong ploidy), B (wrong chromosome count), and D (wrong both). The correct answer is C.

Question 2 (Free Response)

A cell from a diploid organism with $2n=6$ is observed dividing, with 3 chromosomes lined up individually at the metaphase plate. No homologous pairs are present in the cell. (a) Identify the stage and type of cell division this cell is undergoing. Justify your answer. (b) The organism is heterozygous for the seed shape allele $R/r$ on chromosome 1. Assuming no crossing over occurred, list all possible allele arrangements of chromosome 1 in this cell. (c) Calculate the number of distinct combinations of maternal and paternal chromosomes possible in gametes from this organism, and explain how independent assortment contributes to genetic variation.

Worked Solution: (a) This cell is in metaphase of meiosis II. Justification: The organism is diploid with $2n=6$, so diploid cells have 6 total chromosomes. Ploidy is halved in meiosis I, so cells entering meiosis II have 3 haploid chromosomes, which line up individually at the metaphase plate. If this were mitosis or meiosis I, the cell would have 6 chromosomes, so this can only be meiosis II. (b) There are two possible outcomes: either the cell entering meiosis I received the chromosome with $R$, so both sister chromatids have $R$, or it received the chromosome with $r$, so both sister chromatids have $r$. Possible arrangements: 1) Both sister chromatids of chromosome 1 carry $R$; 2) Both sister chromatids of chromosome 1 carry $r$. No crossing over means sister chromatids are identical, so they carry the same allele. (c) For this organism, $n=3$, so the number of combinations is $2^n=2^3=8$. Independent assortment means each homologous pair lines up independently of other pairs, so gametes get random mixes of maternal and paternal chromosomes, rather than all maternal or all paternal chromosomes. This creates far more genetic variation than non-random segregation would.

Question 3 (Application / Real-World Style)

In domestic cats, the diploid number is $2n=38$. A female cat heterozygous for the recessive allele $d$ that causes deafness ($Dd$, where $D$ causes normal hearing) has nondisjunction of chromosome 4 (the chromosome carrying the deafness allele) during meiosis I. The resulting egg is fertilized by a normal sperm from a heterozygous male cat ($Dd$). What is the chromosome number of the resulting zygote, and what is the probability that the kitten will be deaf?

Worked Solution:

1. Normal cat gametes have $n=19$ chromosomes. The egg has an extra copy of chromosome 4 due to nondisjunction, so the egg has $19+1=20$ chromosomes. The sperm is normal, so it has 19 chromosomes.
2. Total zygote chromosome number: $20+19=39$ chromosomes.
3. Nondisjunction in meiosis I of the female means both the $D$ and $d$ homologs end up in the same egg, so the egg genotype is $Dd$. The male can produce either a $D$ or $d$ sperm with equal probability (50% each).
4. Possible zygote genotypes: $DDd$ (50% probability, has at least one dominant $D$, so normal hearing) or $Ddd$ (50% probability, two copies of recessive $d$, so deaf). Probability of deafness is 50% or 1/2.

Interpretation: The resulting kitten will have an extra chromosome 4, and has a 50% chance of being born deaf due to inheriting two copies of the recessive deafness allele.

7. Quick Reference Cheatsheet

8. What's Next

Meiosis is the foundational prerequisite for all concepts in Unit 5 Heredity, and for the study of evolution across the AP Biology course. Next you will apply the patterns of chromosome segregation during meiosis to Mendelian genetics, where you will connect the law of segregation and independent assortment directly to the behavior of homologous chromosomes during meiosis I. Without mastering ploidy changes and segregation in meiosis, you cannot correctly predict inheritance patterns for any cross, or interpret non-Mendelian patterns like linkage, sex-linkage, and polygenic inheritance. Beyond heredity, the genetic variation generated by meiosis is the raw material for natural selection, so it is critical for understanding how populations evolve over time.

• Mendelian Genetics
• Non-Mendelian Inheritance
• Natural Selection
• Population Genetics