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3 years ago

Identify two structural features of purines and Pyrimidines

Chemistry

1 answer:

Elena L [17]3 years ago

5 0

Answer:

The answers are:

Purines:

C. contain four ring nitrogen atoms.

E. contain two heterocyclic rings.

Pyrimidines:

C. contain only two ring nitrogen atoms.

E. contain one heterocyclic ring.

Explanation:

Purines and Pyrimidines are nitrogenous bases which are the building blocks of nucleic acids (DNA and RNA).

Purines are composed by two fused heterocyclic rings, one of them is a 6-ring and the other is a 5-ring. Each ring contains two nitrogen atoms which form part of the ring. Thus, the nitrogen positions in purines are: 1', 3', 7' and 9'. Depending on the functional groups bonded to the two-ring structure, a purine base can be Guanidine (G) or Adenine (A).

The structure of Pyrimidines is a single heterocycle ring wich contains two nitrogen atoms in positions 1' and 3'. Depending of the functional groups, they can be: Cytosine (C), Thymidine (T) and Uracil (U, which is found in RNA).

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Explain how tectonic plate movement could create another supercontinent like pangaea

Ksenya-84 [330]

The plates are moving and are for billions of years... they collide and that they rip apart. The plates carry the continents on their "backs"; if 2 plates -each with a continent- collide, then a supercontinent is created. The landmass of Asian nation accustomed be an isolated block concerning 30-50 million years ago; it 1st crashed into Asia to form the range of mountains. If the movement of the plates brought all the continents along once more, one continent would type. At the instant plate, movements aren't heading in this direction.

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3 years ago

Read 2 more answers

How much of a sample remains after five half-lives have occurred?

timama [110]

Answer:

1/32 of the original sample

Explanation:

We have to use the formula

N/No = (1/2)^t/t1/2

N= amount of radioactive sample left after n number of half lives

No= original amount of radioactive sample present

t= time taken for the amount of radioactive samples to reduce to N

t1/2= half-life of the radioactive sample

We have been told that t= five half lives. This implies that t= 5(t1/2)

N/No = (1/2)^5(t1/2)/t1/2

Note that the ratio of radioactive samples left after time (t) is given by N/No. Hence;

N/No= (1/2)^5

N/No = 1/32

Hence the fraction left is 1/32 of the original sample.

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3 years ago

A system does 125 J of work and cools down by releasing 438 J of heat. The change in internal energy is ____________ J.

Jet001 [13]

Given that,

Work done by the system = 125 J

Energy released when it cools down = 438 J

To find,

The change in internal energy.

Solution,

As heat is released by the system, Q = -438 J

Work done by the system, W = -125 J

Using the first law of thermodynamics. The change in internal energy is given by :

$\Delta U=Q-W\\\\=-438-125\\\\=-563\ J$

So, the change in internal energy is 563 J.

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3 years ago

1, 2 and 3 ........

Orlov [11]

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