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11Alexandr11 [23.1K]
3 years ago
6

When a child's bone is fractured growth may be stimulated at the epiphyseal plate of that bone. whnat problems might this extra

growth cause?
Biology
1 answer:
kozerog [31]3 years ago
3 0
The bone growth in the uninjured limb would pass the injured limbs growth due to the broken limb healing instead of growing. the broken limbs epiphyseal plate would be healing the broken bone instead of growing. the uninjured limb would grow beyond the injured limb causing lack of coordination, unbalance, and its a possibility that it can cause pain to the child's back or hips.. 
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Which of the following statements correctly relates mutations and survival rates of plants?
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The third one is the right answer.
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Organisms made up of one or more cells with complex internal structure are called
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Which parts of photosynthesis occur in the thylakoid membranes of the grana?.
abruzzese [7]

Answer:

The light-dependent reactions take place in the thylakoid membranes in the granum (stack of thylakoids), within the chloroplast. Figure: The two stages of photosynthesis: Photosynthesis takes place in two stages: light-dependent reactions and the Calvin cycle (light-independent reactions).

Explanation:

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6 0
2 years ago
Rock particles dropped by wind or water are called?
ANTONII [103]
Sediments is the answers
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3 years ago
assume that life on Mars requires cell potential to be 100mV, and the extracellular concentrations of the three major species ar
svetlana [45]

Answer:

Explanation:

From the information given:

The cell potential on mars E = + 100 mV

By using Goldman's equation:

E_m = \dfrac{RT}{zF}In \Big (\dfrac{P_K[K^+]_{out}+P_{Na}[Na^+]_{out}+P_{Cl}[Cl^-]_{out} }{P_K[K^+]_{in}+P_{Na}[Na^+]_{in}+ P_{Cl}[Cl^-]_{in}}      \Big )

Let's take a look at the impermeable cell with respect to two species;

and the two species be Na⁺ and Cl⁻

E_m = \dfrac{RT}{zF} In \dfrac{[K^+]_{out}}{[K^+]_{in}}

where;

z = ionic charge on the species = + 1

F = faraday constant

∴

100 \times 10^{-3} = \Big (\dfrac{8.314 \times 298}{1\times 96485} \Big) \mathtt{In}  \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

3.981= \mathtt{In} \Big ( \dfrac{4}{[K^+]_{in}}   \Big)

exp ( 3.981) = \dfrac{4}{[K^+]_{in}} \\ \\  53.57 = \dfrac{4}{[K^+]_{in}}

[K^+]_{in} = \dfrac{4}{53.57}

[K^+]_{in}  = 0.0476

For [Cl⁻]:

100 \times 10^{-3} = -0.0257 \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

-3.981 =  \  \mathtt{In} \Big ( \dfrac{120}{[Cl^-]_{in}}   \Big)

0.01867 =  \dfrac{120}{[Cl^-]_{in}}

[Cl^-]_{in} = \dfrac{120}{0.01867}

[Cl^-]_{in} =6427.4

For [Na⁺]:

100 \times 10^{-3} = 0.0257 \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

53.57= \Big ( \dfrac{145}{[Na^+]_{in}}   \Big)

[Na^+]_{in}= 2.70

6 0
3 years ago
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