A mutation occurs in a germ cell of a pure-breeding, wild-type male mouse prior to DNA replication. The mutation is not corrected, and the cell undergoes DNA replication and a normal meiosis produces four gametes. How many of these gametes will carry the mutation?

Answer: 2

Explanation:

A germ cell is a cell that gives rise to gametes. This is through meiosis, a type of cell division that reduces the chromosome number by half. In this way, four haploid cells are created, each genetically different from the parent cell.

During meiosis I, crossing over takes place. It consists on the exchange of genetic material between two homologous chromosomes to create recombinant chromosomes. Then, if there is a mutation not corrected, it would end up only in one of those two chromosomes. At the end of meiosis I, these homologous chromosomes are separated into two daughter cells. During meiosis I, the sister chromatids of each chromosome within the two daughter cells are separated, creating four daughter gametes.

So, the mutation that was only on one chromosome ended up in one of the two cells created in meiosis I. Then that cell with the mutation divided again during meiosis II, creating two more cells. So, here we have two gametes with one mutation each. On the other hand, the cell that did not receive the mutation also created two daughter cells in meiosis I but with their normal genes. And that same one created two other cells during meiosis II, also normal. The total is 4 cells, only 2 affected by the mutation.