ở gà có bộ nhiễm sắc thể 2n=78 xét hai tế bào sinh dục cơ sở khai phương ở vùng sinh sản đều nguyên phân liên tiếp 4 đợt 25% tế bào con trải qua giảm phân xác định số nst đơn môi trường cần phải cung c...

The question is asking for the number of single chromatids (NST) that need to be provided in the environment for the process of reduction division. From the given information, we know that there are two germ cells undergoing consecutive meiotic divisions for 4 rounds. And 25% of the resulting cells undergo reduction division. To calculate the number of NST, we need to consider the number of cells produced after each round of meiotic division. In the first round, each germ cell produces 2^1 = 2 cells. In the second round, each cell from the first round produces 2^2 = 4 cells. In the third round, each cell from the second round produces 2^3 = 8 cells. In the fourth round, each cell from the third round produces 2^4 = 16 cells. So, after 4 rounds of meiotic division, each germ cell will produce a total of 2 x 4 x 8 x 16 = 1024 cells. Since 25% of these cells undergo reduction division, the number of cells undergoing reduction division is 1024 x 0.25 = 256 cells. Therefore, the number of single chromatids (NST) that need to be provided in the environment for the process of reduction division is 256.