TritionalshiftsofEscherichiacoliB/rtorichermediahavebeenanalyzedinsynchronouslygrowingandexponential-phasepopulations.Earlyperturbationsinthetimingofcelldivisionwereobserved.Attheslowgrowth,divisionpro-gressedatarateequaltoorlessthanthepreshiftrateforabout1h.Atintermediategrowth,bothdelaysandaccelerationindivisionwereobserved.Theextentoftheperturbationdependedupontheageofthecellsatthetimeoftheshiftandthecompositionofthepreshiftandpostshiftmedia.TheperturbationwasdifferentinthetwosubstrainsofE.coliB/r I got this from http://jb.asm.org/content/136/2/631.full.pdf hopefully it helps you
There are all sorts of ways to reconstruct the history of life on Earth. Pinning down when specific events occurred is often tricky, though. For this, biologists depend mainly on dating the rocks in which fossils are found, and by looking at the “molecular clocks” in the DNA of living organisms.
There are problems with each of these methods. The fossil record is like a movie with most of the frames cut out. Because it is so incomplete, it can be difficult to establish exactly when particular evolutionary changes happened.
Modern genetics allows scientists to measure how different species are from each other at a molecular level, and thus to estimate how much time has passed since a single lineage split into different species. Confounding factors rack up for species that are very distantly related, making the earlier dates more uncertain.
These difficulties mean that the dates in the timeline should be taken as approximate. As a general rule, they become more uncertain the further back along the geological timescale we look. Dates that are very uncertain are marked with a question mark.
Answer:
The genotypes found in the F2 offspring will be An1An1, An1An2 and An2An2.
Explanation:
In the example given in the question, the color of the petals of snapdragon flowers are given with the An1 allele having full activity and the An2 allele being the null allele. For the F1 offspring, two snapdragon flowers are crossed, one with red petals which is An1An1 and one with ivory petals which is An2An2. All the genotypes for the F1 offspring will result in An1An2.
Then using this F1 offspring, the F2 offspring is found and the genotype of the F2 offspring will be An1An1 / An1An2 / An1An2 / An2An2.
So the genotypes found in the F2 offspring will be An1An1, An1An2 and An2An2. One red petal snapdragon, 1 ivory snapdragon and 2 pink snapdragons.
I hope this answer helps.
Answer: 48.93 mL of sprite
Volume of blood in 7 Kg human = 5 L
Percentage of plasma in blood = 55%
Volume of plasma in 5 L blood = (1 L = 10 dL)
Concentration of glucose in plasma = 80 mg/dL
Amount of glucose present in 27.5 dL : 80 mg /dL × 27.5 dL= 2200 mg
Let the volume of sprite with 2200 mg glucose be x
Concentration of glucose in sprite = 44.96 mg/mL
Explanation:
Answer:
The answer is sponge
Hope this helped mark BRAINLEST!!!
