Option (A) identifies them correctly.
During Meosis 2 daughter cells are formed from a single cell and the chromosomes are equally divided in both the cells.
So, Cell P had 60 chromosome, after miosis 2 daughter cells of each having 30 chromosomes formed, and again after Miosis of from each of those 2 those daughter cells, 2 more daughter cells are formed with equal naumber of chtomosome which equals 15.
D
Just double the amount of bacteria.
1024 * 2 = 2048
Answer:
The order of genes is- dp-cl-ap
Explanation:
The genes a be mapped on chromosome on the basis of the recombination frequency as the recombination shows that is the genes are linked or not. In genetic mapping, 1 per cent recombination frequency is considered as 1 cM or map distance.
In the given question, the three gene distance are provided on the basis of which the gene can be mapped in the following manner:
1. Take the higher value of gene distance that is 42 m.u and plot them on the line,
2. now place second-highest value on the line that is 39 m.u.
3. The gene will be arranged in the order of dp-cl-ap.
4. dp--³---cl-----------³⁹---------------------ap.
Thus, dp-cl-ap is correct.
The red-legged frog breeds in ephemeral ponds from January through March. Its relative, the bullfrog, breeds in permanent ponds from late March through May. There are two pre-mating isolation mechanism at play here,
1. Ecological isolation: The habitats are different, hence the individuals of both the species do not meet. Ephemeral ponds are temporary ponds that develop during rainy days, while the permanent ponds are full of water throughout the year.
2. Temporal isolation: The time of mating is different for both the species to avoid contact between the individuals of closely related species. Red-legged frogs mate in January to March slot and the bullfrog in the March to May slot.
Answer:
they have influenced it through genetic engineering
Explanation: