B is the answer igneous Rock weathering and erosion burail deposition is the answer tqq
Answer:
Due to less steps and requires less energy.
Explanation:
The bacterial cell is able to use glucose first as an energy source then switch to lactose because glucose requires less steps and less amount of energy for the break down as compared to lactose. If lactose is the only sugar available to the bacterial cells, then bacterial cells will use it as energy source for the production of energy. In order to use lactose, the bacteria must express the lac operon genes, which encode the main enzymes for lactose uptake and metabolism.
Answer
I added 4 and 5 as an attachment due to time constraint.
Explanation:
1. at a steady state ca2+ taken by vesicle = 40n/mol
enclosed volume of CSR = 5μ/mg
concentration of ca2+ in vesicle =
40*10⁻⁶/5x10⁻⁶
= 0.008
= 8x10⁻⁹mol/L
2. radius of circle =75nM
volume = 150nm
volume of vesicle = 4/3 x pi x r³
= 1.33 x 3.14 x (75x10⁻⁹)³
= 1.77 x 10⁻²¹
when we convert this we get
1.77x10⁻¹⁸L since 1m³ is equal to 100oL
surface area =4x3.14x(75x10⁻⁹)²
= 12.56x5.625⁻¹⁵
= 7.065x10⁻¹⁴m²
this is 7.065x10⁻¹⁰cm²
3. number of vesicle/CSR
= 5x10⁻⁶/1.77x10⁻¹⁸
= 2.824x10¹²
surface are = 2.824x10¹² x7.06x10⁻¹⁰
= 19.95x10²
Synapsis:
•Early
during the first nuclear division.
•Homologous
chromosomes pair along their length.
•Held
tightly by protein ‘zipper’.
<span>Homologous Recombination: </span>
•Genetic
exchange (crossing
over<span>)
occurs between homologous chromosomes. It then becomes a mix of both paternal and maternal genes (father and mother respectively)</span>
Answer:
In order to find average speed during each interval, we need to divide the distance during those intervals with the period of time. So, for the first interval (day 0 to day 2) hawksbill started from 0 and reached 10 kilometers by the end of the second day. That means that it crossed 10 kilometers in 2 days, so the average speed is 10/2 which is 5 km/day. Similarly, we can calculate speed for other intervals:
• day 2 - day 3: it went from 10 to 12 km in one day, which means it crossed 2 km in one day, so the average speed is 2/1 = 2 km/day
• day 3 - day 4: at the end of the third day it reached 12 km and at the end of the day 4 it remained at 12 km. That means the hawksbill wasn't moving in that interval so the speed was 0
• day 4 - day 5: it went from 12 km to 18 km, which means it crossed 18-12=6 km in one day, so the average speed is 6/1=6 km/day
• day 5 - day 6: it went from 18 to 24 km, which means it crossed 24-18=6 km in one day, so the speed was 6/1=6 km/day
So, to summarize, during the first interval turtle was moving with average speed of 5 km/day, then 2 km/day, in the third interval it wasn't moving and in the last two intervals, it moved in average speed of 6 km/day.
