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Genetics and Evolution (AHL) - Meiosis

Grade 12IBBiology

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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During Prophase I, homologous chromosomes undergo synapsis to form bivalents (tetrads). Crossing over occurs between non-sister chromatids at points called chiasmata, leading to the exchange of genetic material.

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Crossing over produces new combinations of alleles on the chromosomes of the haploid cells, resulting in recombinant gametes that increase genetic diversity.

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Independent assortment of genes is due to the random orientation of pairs of homologous chromosomes (bivalents) in Metaphase I. The direction in which one bivalent faces is independent of the orientation of other bivalents.

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The number of possible combinations of chromosomes produced by independent assortment is 2n2^n, where nn is the haploid number. In humans, this is 2232^{23} (over 8 million combinations), excluding the effects of crossing over.

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Meiosis I is a reduction division (2n→n2n \rightarrow n) where homologous chromosomes are separated. Meiosis II is similar to mitosis (n→nn \rightarrow n), where sister chromatids are separated into different daughter cells.

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Non-disjunction is the failure of chromosomes to separate properly during Anaphase I or Anaphase II. This leads to aneuploidy, such as Down Syndrome, which is characterized by Trisomy 21Trisomy\ 21 (three copies of chromosome 21).

📐Formulae

Possible Chromosome Combinations=2n\text{Possible Chromosome Combinations} = 2^n

Recombination Frequency=Number of RecombinantsTotal Number of Offspring×100\text{Recombination Frequency} = \frac{\text{Number of Recombinants}}{\text{Total Number of Offspring}} \times 100

Diploid Number=2n\text{Diploid Number} = 2n

Haploid Number=n\text{Haploid Number} = n

💡Examples

Problem 1:

A cell from an organism with a diploid number of 2n=122n = 12 undergoes meiosis. Calculate the number of possible genetic combinations in the gametes due to independent assortment alone.

Solution:

26=642^6 = 64 combinations.

Explanation:

The haploid number nn is half of the diploid number 2n2n. Since 2n=122n = 12, then n=6n = 6. Using the formula for independent assortment, 2n=26=642^n = 2^6 = 64.

Problem 2:

In a test cross involving linked genes, 4040 offspring show parental phenotypes and 1010 offspring show recombinant phenotypes. Calculate the recombination frequency.

Solution:

1040+10×100=20%\frac{10}{40 + 10} \times 100 = 20\%

Explanation:

Recombination frequency is calculated by dividing the number of recombinant offspring by the total number of offspring (parental + recombinants) and multiplying by 100100.

Problem 3:

During which phase of meiosis do chiasmata become visible, and what process do they signify?

Solution:

Chiasmata become visible during Prophase I; they signify the completion of crossing over.

Explanation:

In Prophase I, homologous chromosomes pair up (synapsis). Crossing over occurs when non-sister chromatids break and rejoin at the chiasmata, exchanging DNA segments.

Meiosis - Revision Notes & Key Diagrams | IB Grade 12 Biology