Spreading black necrosis, swelling, pain and froth or bubbles are characteristic of

Both human and bacteria cells contain DNA and both have a outer membrane that controls how materials come into and leave the cell.

Predominantly multicellular not in bacteria but in human body

*Cell contains a nucleus and other membrane bound organelles not in bacteria but in human body..

*DNA occurs in a circular form* in bacteria only..

* Ribosomes size in bacteria 70s and in human body 80s

*Capable of growth at temperatures greater than 80 C only in bacteria

*His tone proteins present in cell only in human body cells

* Operons present in DNA only in bacteria..

5 0

1 year ago

During process 1→3→6, the temperature of the gas __________. during process , the temperature of the gas __________. decreases a

kogti [31]

<span>During process 1â†’3â†’6, the temperature of the gas decreases slowly during process , the temperature of the gas rapidly decreases and then increases increases and then decreases remains constant. Hence, in the first step, the temperature of gas decreases slowly, then decreases rapidly, then it increases and finally the decreases becomes constant.</span>

8 0

1 year ago

a 24 kg rock containing 3,528 j of potential energy rests upon the edge of a cliff how tall is the cliff

GenaCL600 [577]

Potential gravitational energy is computed with the formula:

$PEgrav=mgh$

Where:
PEgrav = Potential gravitation energy (J)
m = mass (kg)
h = height (m)
g = acceleration due to gravity (9.8m/s^2)

Take note that acceleration due to gravity is a constant.

All you need to do is put in what you know and solve for what you don't know. So your given is the following:
PEgrav = 3,528J
m = 24 kg
g = 9.8 m/s^2
Put that into your formula:
$PEgrav=mgh$
$3,528J = (24kg)(9.8m/s^{2})(h)$
$3,528J = (235.2kg.m/s^{2})(h)$
$\frac{3,528J}{235.2kg.m/s^{2}} = h$
$15m=h$

The height of the cliff is 15m.

4 0

1 year ago

In this example, we used lotka-volterra equations to model populations of rabbits and wolves. let's modify those equations as fo

Daniel [21]

Answer:

Wolves absence would lead the rabbit population to become stable at 5,000.

Explanation:

Step 1. Concept

If there is a missing variable that should be in the system, set that particular variable to zero and make the necessary conclusion.

Step 2. Given,

dR/dt = 0.08R(1 - 000.2R) - 0.001RW

dW/dt = −0.02W + 0.00002RW

Step 3. Calculation

The transformed equations of populations of rabbit and wolves will be

dR/dt = 0.08R(1 - 000.2R) - 0.001RW

dW/dt = −0.02W + 0.00002RW

Let's check the rabbit population in the absence of wolves(W) to 0

Thus,

Differential equation dR/dt

dR/dt = 0.08R(1 - 000.2R) - 0.001RW

Take W to equal to O

dR/dt = 0.08R(1 - 000.2R) - 0.001R(0)

dR/dt = 0.08R(1 - 000.2R)

When,

dR/dt = 0 we have R = 0 or R = 5,000

This,

R = 5,000 becomes the equilibrium point in the absence of wolves

Then,

0 < R < 5,000, dR/dt > 0

For R, let's estimate a value rise of the rabbit population to 5,000

Then,

R > 5,000, dR/dt < 0

Judging from the value if R, we infer a deduction in the rabbit population by 5,000

Thus,

We can conclude that wolves absence would lead the rabbit population to become stable at 5,000.

7 0

1 year ago

Enolase is an enzyme that catalyzes one reaction in glycolysis in all organisms that carry out this process. The amino acid sequ

lorasvet [3.4K]

Answer:

a) The response indicates that a pH below or above this range will most likely cause enolase to denature/change its shape and be less efficient or unable to catalyze the reaction.

b)The response indicates that the appropriate negative control is to measure the reaction rate (at the varying substrate concentrations) without any enzyme present.

c)The response indicated that the enolase has a more stable/functional/correct/normal protein structure at the higher temperature of 55°C than at 37°C because the enzyme is from an organism that is adapted to growth at 55°C.

Explanation:

Enolase catalyzes the conversion of 2-phosphoglycerate to phosphoenolpyruvate during both glycolysis and gluconeogenesis.In bacteria, enolases are highly conserved enzymes and commonly exist as homodimers.

The temperature optimum for enolase catalysis was 80°C, close to the measured thermal stability of the protein which was determined to be 75°C, while the pH optimum for enzyme activity was 6.5. The specific activities of purified enolase determined at 25 and 80°C were 147 and 300 U mg−1 of protein, respectively. Km values for the 2-phosphoglycerate/phosphoenolpyruvate reaction determined at 25 and 80°C were 0.16 and 0.03 mM, respectively. The Km values for Mg2+ binding at these temperatures were 2.5 and 1.9 mM, respectively.

Enolase-1 from Chloroflexus aurantiacus (EnoCa), a thermophilic green non-sulfur bacterium that grows photosynthetically under anaerobic conditions. The biochemical and structural properties of enolase from C. aurantiacus are consistent with this being thermally adapted.

8 0

1 year ago