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Genetics - Meiosis

Grade 12IBBiology

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

🔑Concepts

Meiosis is a reduction division of a diploid nucleus (2n2n) to form four haploid nuclei (nn).

DNA replication occurs during the SS phase of interphase prior to meiosis, resulting in chromosomes consisting of two identical sister chromatids.

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

Random orientation of bivalents occurs during Metaphase I, where the orientation of one pair of chromosomes is independent of other pairs. This leads to independent assortment.

In Meiosis I, homologous chromosomes are separated, reducing the chromosome number by half. In Meiosis II, sister chromatids are separated.

Genetic variation is promoted by: 1. Crossing over in Prophase I; 2. Random orientation in Metaphase I; 3. Random fusion of gametes during fertilization.

Nondisjunction is the failure of chromosomes to separate properly during Anaphase I or II, which can result in gametes with an extra or missing chromosome, such as in Down Syndrome (Trisomy 2121).

📐Formulae

2n2^n

n=haploid number of chromosomesn = \text{haploid number of chromosomes}

2n=diploid number of chromosomes2n = \text{diploid number of chromosomes}

💡Examples

Problem 1:

A cell from an organism with a diploid number of 2n=122n = 12 undergoes meiosis. Calculate the number of possible chromosome combinations that can be produced in the gametes due to independent assortment, excluding the effects of crossing over.

Solution:

26=642^6 = 64

Explanation:

The formula for the number of possible combinations due to independent assortment is 2n2^n. Given 2n=122n = 12, we find that n=6n = 6. Substituting this into the formula gives 262^6, which equals 6464 possible genetic combinations.

Problem 2:

Explain the chromosome count in a human cell following Meiosis I and Meiosis II, starting from a primary spermatocyte (2n=462n = 46).

Solution:

After Meiosis I: n=23n = 23 chromosomes (each with 22 chromatids). After Meiosis II: n=23n = 23 chromosomes (each with 11 chromatid).

Explanation:

Meiosis I is the reduction division where homologous pairs separate, so the number of chromosomes drops from 2n2n (4646) to nn (2323). Meiosis II is an equational division where sister chromatids separate, so the chromosome count remains n=23n = 23 in each of the four daughter cells.

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