The choices can be found elsewhere and as follows:
A. mature leaves
B. shoot apical meristem
C. cell elongation zone
<span>D. axillary buds
</span>
I think the correct answers are option B and D. It would be at the shoot apical meristem and the axillary buds that <span> a vascular plant would you expect to find totipotent cells. Hope this answers the question.</span>
The answer is Na + is entering the cell and the K + is leaving, during the depolarization phase of the action potential open Na + channels allow Na + ions to diffuse into the cell. This inward movement of positive charge makes the membrane potential more positive. The depolarization phase is a positive response sequence where open Na + channels cause depolarization which in re-occurrence causes more voltage gated Na+ channels to open.
This membrane mimics the plasma membrane that is around animal cells. In order to explore how water moves in and out of a cell, one cell was weighed and then submerged in hypertonic solution. The other egg was weighted and submerged in hypotonic solution. It was predicted that the egg submerged in hypertonic solution would decrease in mass. It was predicted that the egg submerged in hypotonic solution would increase in mass.
Explanation:
The outermost covering of an animal cell is the plasma membrane. It is a selectively permeable membrane that allows only selective molecules to pass through it.
A solution having higher concentration of solute than the cell cytoplasm is called a hypertonic solution.
A solution having lower concentration of solute than the cell cytoplasm is called the hypotonic solution.
The movement of water molecules from the region of its higher concentration to the region of its lower concentration through a semipermeable membrane is called osmosis.
A cell placed in hypertonic solution will undergo exosmosis so it will lose water and its mass will decrease.
A cell placed in hypotonic solution will undergo endosmosis of water so it will gain water and its mass will increase.
Answer:
- 1/2 Aa Wx wx (colored seeds, normal starch)
- 1/2 aa Wx wx (colorless seed, normal starch).
Explanation:
- A (colored seed) is dominant over a (colorless)
- Wx (normal starch) is dominant over wx (waxy)
Both loci are independent.
A <em>Aa WxWx </em>individual was test crossed (crossed with a homozygous recessive <em>aa wxwx</em> individual).
- The homozygous recessive can only produce <em>a wx </em>gametes.
- The dihybrid individual can produce two types of gametes, all of them with the same frequency because the genes segregate independently: 1/2 <em>A Wx </em>and 1/2<em> a Wx </em>
<u>The possible offspring resulting from the combination of those gametes is:</u>
- 1/2 Aa Wx wx (colored seeds, normal starch)
- 1/2 aa Wx wx (colorless seed, normal starch).
answer is the first1 off spring to parants
