Ask question

Ask question

All categories

3 years ago

8

Match the specific enzyme to its class. A. oxidoreductase B. hydrolase C. transferase D. ligase E. lyase aminase

1 answer:

lilavasa [31]3 years ago

3 0

Answer: Every enzyme has a specific name that can give us insight into the specific reaction that that enzyme can catalyze. We divide them into six different categories.

1) Oxidoreductase - includes two different types of reactions by transferring electrons from either molecule A to B or vice versa. It is involved in oxidizing electrons away from a molecule.

2) Hydrolase - uses water to divide a molecule into two other molecules.

3) Transferase - you move some functional group X from molecule B to molecule A

4) Ligase - catalyzes reactions between two molecules, A and B, that are combining to form a complex between the two. (example: DNA replication)

5) Lyase - divides a molecule into two other molecules without using water and without reducing or oxidation

You might be interested in

The melting point of aluminum is about 660 °C. How many degrees above the boiling point of water is this temperature?

Arte-miy333 [17]

560 degrees above water

8 0

3 years ago

A powdered drink mix is stirred into a glass of water to form a solution. If any more is added it will not dissolve

Brut [27]

SaturTED, hope this helps, m8

6 0

3 years ago

Suppose a 1.0 L solution contains 0.020 M in Cu(NO3)2, then 0.40 moles of NH3 are added. Assuming no change in volume, what is t

Rasek [7]

Answer:

$4.6*10^{-14}$ M

Explanation:

Concentration of $Cu^{2+} = [Cu(NO_3)_2]$ = 0.020 M

Constructing an ICE table;we have:

$Cu^{2+}+4NH_3_{aq} \rightleftharpoons [Cu(NH_3)_4]^{2+}_{(aq)}$

Initial (M) 0.020 0.40 0

Change (M) - x - 4 x x

Equilibrium (M) 0.020 -x 0.40 - 4 x x

Given that: $K_f =1.7*10^{13}$

$K_f } = \frac{[Cu(NH_3)_4]^{2+}}{[Cu^{2+}][NH_3]^4}$

$1.7*10^{13} = \frac{x}{(0.020-x)(0.40-4x)^4}$

Since x is so small; 0.40 -4x = 0.40

Then:

$1.7*10^{13} = \frac{x}{(0.020-x)(0.0256)}$

$1.7*10^{13} = \frac{x}{(5.12*10^{-4}-0.0256x)}$

$1.7*10^{13}(5.12*10^{-4} - 0.0256x) = x$

$8.704*10^9-4.352*10^{11}x =x$

$8.704*10^9 = 4.352*10^{11}x$

$x = \frac{8.704*10^9}{4.352*10^{11}}$

$x = 0.0199999999999540$

$Cu^{2+}= 0.020 - 0.019999999999954$

$Cu^{2+} = 4.6*10^{-14}$ M

8 0

3 years ago

The main environmental factors that affect microorganism growth include chemical inhibitors, oxygen, pH, nutrients, moisture and

aev [14]

The main environmental factors that affect microorganism growth include chemical inhibitors, oxygen, pH, nutrients, moisture and temperature.

<h3>What are micro organism?</h3>

Micro Organism are organism that can be seen with a microscope.

Microorganisms include bacteria, protozoa, algae, and fungi.

Micro organisms exist in its single-celled form or as a colony of cells.

The main environmental factors that affect microorganism growth include chemical inhibitors, oxygen, pH, nutrients, moisture and temperature.

learn more on micro organism here: brainly.com/question/18051752

#SPJ12

3 0

2 years ago

Ifa dozen nails weighs 1 pound, how many nails are in 7.6 pounds?

NikAS [45]

Then there should be mostly 91 nails

4 0

3 years ago