If two Bb guinea pigs cross, there are four possible outcomes:
- both pass the B: the son is BB
- one of the pass the B and the other pass the b: the son is Bb
- both pass the b: the son is bb
Since B (being black) is dominant, the son will be black in both BB and Bb cases. In order for a son to be brown, he must be a pure bb specimen.
So, if all of the offspring are black, it means that all four guinea pigs are either BB or Bb, which in turn means that at least one of the parents passed the B gene.
Answer:
The correct option is C - Professor Scrawll applied 100 nM BurD to the cells for 24 hours, while Dr. Bogey applied 1 nM BurD to the cells for 12 hours.
Explanation:
The correct option is C - Professor Scrawll applied 100 nM BurD to the cells for 24 hours, while Dr. Bogey applied 1 nM BurD to the cells for 12 hours.
As the exotoxin, BurD is very stable and lyse the ankle cells very quickly, more concentration and more time of action should only lyse the cells. Perhaps Dr.Bogey's ankle cells were not lysed because the concentration she used was only 1nM compared to the 100nM concentration used by Dr. Scrawll, and the time period of incubation was only 12 hours compared to the 24 hours used by Dr. Scrawll.
Considering the other explanations given in the remaining options, the concentration and time of incubation used by Dr. Bogey are more than that used by Dr. Scrawll which should only have possibly lysed the cells. Moreover, contamination with bleach also should have only lysed the cells.
Answer:
8
Explanation:
The fly has only 3 pairs of somatic chromosomes, that means that only can combine 3 chromosomes. To calculate the combinations the exponential 2n it is used. In this case is 2 to the third power that is equal = 2 x 2 x 2 = 8
This fly also has a fourth pair, but this one is the sexual one, so the main function of this gene is to give the sex to the offspring.
Answer: C, Fungus X is a common mold and fungus Y is a Chytrid.
Explanation:
I just took the test
awnser is high concentrations of oxygen
