Answer:
In a fruit fly experiment, grey body, normal winged (homozygous dominant) fruit flies were mated with black body, short winged (homozygous recessive) fruit flies. The F1 dihybrid females were then used in a test cross. If the genes are always linked and no crossing over occurs, what would be the predicted ratio in the F2 generation?
GG x bb = Gb, Gb, Gb and Gb F1 generation
grey body heterozygous offspring 4:0
Gb x Gb= GG, Gb, Gb, and bb F2 generation
3:1 three grey body fly and one black body fly
Explanation:
No. all lipids r semi-solid, like room temp. butter..
Answer:
(08) and (32)
Explanation:
To make osmosis happen there has to be a difference in the concentration of solutes, between the inside and the outside of the cell. To valance this difference in concentration, water has to flow towards a place that has a higher number of solutes.
The lipids in the cell's wall make this membrane semipermeable. This allows the passage of specific components only, such as water through aquaporins. Lipids and other elements are of importance in the barrier because they maintain the cell separated from the outside, allowing it to be balanced as regards the different substances that can interact with it.
This would be 7. Below 7 is acidic and above is basic. 7 is neutral
Answer:
a. movement can occur both upward and downward in the plant
Explanation:
The phloem loading causes the accumulation of sugars in the sieved elements generating a negative solute potential (quedas), with a drop in water potential (ψw), so water enters the sieved elements increasing the turgor pressure (ψp). With the discharge of phloem in the drain occurs lower concentration of sugars in the screened elements, increases the solute potential, becoming positive, thus the phloem water potential increases and thus the water leaves the conducting vessel. In the specific case of sugar movement in the phloem, it can be stated that this movement can occur both up and down in the plant.
