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andriy [413]
3 years ago
8

What is the relationship between the sequence of DNA’s subcomponents and the properties of DNA? 1. The amino acid sequence along

a segment of DNA determines the three-dimensional shape of the protein that will be produced. 2. The deoxyribose sequence in a DNA molecule determines the complementary molecule that will form double stranded DNA. 3. The nitrogenous base sequence along a segment of DNA determines the amino acid sequence that will be produced. 4. The sequence of enzymes in a DNA molecule determines the direction in which DNA will be transcribed into mRNA.
Chemistry
1 answer:
wel3 years ago
7 0

Answer:

3.

Explanation:

There are four nitrogenous bases in a DNA including guanine (G), adenine (A), cytosine (C) and Thymine (T) and together they form the nitrogenous base sequence arranged in a specific order of three letters such as GAC and TAG to form a genetic code.

These nitrogenous base sequences forming genetic code are amino acid specific and determine the amino acid sequence in DNA. for example: CTT determines leucine and GTT determines valine.

Hence, the correct option is "3".

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5 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
The radius of a vanadium atom is 130 pm. How many vanadium atoms would have to be laid side by side to span a distance of 1.30 m
monitta

Answer:

5 000 000 (5 million atoms)

Explanation:

Let us assume that a vanadium atom has a spherical shape.

diameter of a sphere = 2 x radius of the sphere

Thus,

Radius of a vanadium atom = 130 pm

                                              = 130 x 10^{-12} m

The diameter of a vanadium atom = 2 x radius

                                                         = 2 x 130 x 10^{-12}

                                               = 260 x 10^{-12} m

Given a distance of 1.30 mm = 1.30 x 10^{-3} m,

The number of vanadium atoms required to span the distance = \frac{1.3*10^{-3} }{260*10^{-12} }

                                                  = 5000000

Therefore, the number of vanadium atom that would span a distance of 1.30 mm is 5 million.

3 0
3 years ago
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