X-intercept =2; y-intercept =4/3 Find he equation of the line

DNA molecules consist of chemically linked sequences of the bases adenine, guanine, cytosine and thymine, denoted A, G, C and T.

Dmitry [639]

Answer:

1. See the attached tree diagram (64 different sequences); 2. 64 codons; 3. 8 codons; 4. 24 codons consist of three different bases.

Step-by-step explanation:

The main thing to solve this kind of problem, it is to know if the pool of elements admits repetition and if the order matters in the sequences or collections of objects that we can form.

In this problem, we have the bases of the DNA molecule, namely, adenine (A), thymine (T), guanine (G) and cytosine (C) and they may appear in a sequence of three bases (codon) more than once. In other words, repetition is allowed.

We can also notice that order matters in this problem since the position of the base in the sequence makes a difference in it, i.e. a codon (ATA) is different from codon (TAA) or (AAT).

Then, we are in front of sequences that admit repetitions and the order they may appear makes a difference on them, and the formula for this is as follows:

$\\ Sequences\;with\;repetition = n^{k}$ (1)

They are sequences of k objects from a pool of n objects where the order they may appear matters and can appeared more than once (repetition allowed).

<h3>1 and 2. Possible base sequences using tree diagram and number of possible codons</h3>

Having all the previous information, we can solve this question as follows:

All possible base sequences are represented in the first graph below (left graph) and are 64 since n = 4 and k = 3.

$\\ Sequences\;with\;repetition = 4^{3} = 4*4*4 = 64$

Looking at the graph there are 4 bases * 4 bases * 4 bases and they form 64 possible sequences of three bases or codons. So there are 64 different codons. Graphically, AAA is the first case, then AAT, the second case, and so on until complete all possible sequences. The second graph shows another method using a kind of matrices with the same results.

<h3>3. Cases for codons whose first and third bases are purines and whose second base is a pyrimidine</h3>

In this case, we also have sequences with repetitions and the order matters.

So we can use the same formula (1) as before, taking into account that we need to form sequences of one object for each place (we admit only a Purine) from a pool of two objects (we have two Purines: A and G) for the first place of the codon. The third place of the codon follows the same rules to be formed.

For the second place of the codon, we have a similar case: we have two Pyrimidines (C and T) and we need to form sequences of one object for this second place in the codon.

Thus, mathematically:

$\\ Sequences\;purine\;pyrimidine\;purine = n^{k}*n^{k}*n^{k} = 2^{1}*2^{1}*2^{1} = 8$

All these sequences can be seen in the first graph (left graph) representing dots. They are:

$\\ \{ATA, ATG, ACA, ACG, GTA, GTG, GCA, GCG\}$

The second graph also shows these sequences (right graph).

<h3>4. Possible codons that consist of three different bases</h3>

In this case, we have different conditions: still, order matters but no repetition is allowed since the codons must consist of three different bases.

This is a case of permutation, and the formula for this is as follows:

$\\ nP_{k} = \frac{n!}{n-k}!$ (2)

Where n! is the symbol for factorial of number n.

In words, we need to form different sequences (order matters with no repetition) of three objects (a codon) (k = 3) from a pool of four objects (n = 4) (four bases: A, T, G, and C).

Then, the possible number of codons that consist of three different bases--using formula (2)--is:

$\\ 4P_{3} = \frac{4!}{4-3}! = \frac{4!}{1!} = \frac{4!}{1} = 4! = 4*3*2*1 = 24$

Thus, there are 24 possible cases for codons that consist of three different bases and are graphically displayed in both graphs (as an asterisk symbol for left graph and closed in circles in right graph).

These sequences are:

{ATG, ATC, AGT, AGC, ACT, ACG, TAG, TAC, TGA, TGC, TCA, TCG, GAT, GAC, GTA, GTC, GCA, GCT, CAT, CAG, CTA, CTG, CGA, CGT}

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6 0

3 years ago

Mr. Smith built a rectangular-shaped sandbox for his children, measuring 4 feet by 4 feet by 6 inches. Sand has a density of app

ehidna [41]

Answer:

To completely fill the sandbox will cost $56

Step-by-step explanation:

First, we calculate volume of the rectangular-shaped sandbox:

$V=long*with*height\\V=4feet*4feet*0.5feet\\V=8cubicfoot$

We know that density of sand is 100 pounds per cubic foot. Then, to fill the sand box that has a volume of 8 cubic foot, we calculate by rule 3:

1cubic foot..............100 pounds

8 cubic foot.............x pounds

$x=\frac{8*100}{1}$

$x=800 pounds$

We will need 800 pounds of sand to fill the sandbox, then we can calculate by rule 3 the number of bags needed and next the total cost of them:

50 pounds.......1bag

800 pounds.....x bags

$x=\frac{800*1}{50} \\x=16$

To calculate the cost:

1 bag ..................$3.50

16 bags.................$x

$x=16*3.50\\x=$56$

To completely fill the sandbox will cost $56

4 0

3 years ago

In the equation, y = 0.007x, how many inches of rain falls in 6 minutes?

djverab [1.8K]

Answer:

X represents the value you can multiply with the slope to get the answer.

y = 0.007x

= 0.007(6)

= 0.042

Therefore, 0.042 inches of rain falls in 6 minutes.

7 0

3 years ago

Is 18/54 equivalent to 3/9?

viva [34]

Yessssssssssssssssssssssss

4 0

3 years ago

Please help. I have attached a picture.

Ksivusya [100]

Answer: Choice C) -1/3

The given line has a slope of -1/3 so anything parallel to that given line is also going to have an equal slope (of -1/3)

To find the slope of that line shown, notice how we drop down 1 unit and the move to the right 3 units when we go from the left point to the right point. Alternatively, you can use the slope formula m = (y2-y1)/(x2-x1) to get the same result.

7 0

3 years ago

X-intercept =2; y-intercept =4/3 Find he equation of the line​

X-intercept =2; y-intercept =4/3 Find he equation of the line