What mRNA sequence would result from the following DNA sequence?

Physics

1 answer:

Nezavi [6.7K]3 years ago

5 0

Answer:

UAC CUG AGG AUC

Explanation:

The mRNA sequence from ATG GAC TCC TAG DNA sequence would be UAC CUG AGG AUC.

According to Chargaff's base pairing rule, the purine bases always pair with pyrimidine bases. Specifically, Adenine base must pair with Thymine base while Guanine base must pair with Cytosine base. In RNA, Thymine base is replaced with Uracil base.

Hence:

ATG GAC TCC TAG will pair with

UAC CUG AGG AUC

You might be interested in

When placed in a ________ solution, a cell will exhibit a net loss of water. the specific process by which the water leaves the

katen-ka-za [31]

The specific process by which the water leaves the cell is called plasmolysis.

<h3>What is plasmolysis?</h3>

Plasmolysis is the process in which the cell loses the water that is present inside it because the environment of the cells produces pressure to build up inside the cell. If a cell is present in a hypertonic solution means that solvent concentration is lower in the surrounding solution, then water comes out of the cell while on the other hand, if a cell is present in a hypotonic solution in which the solvent concentration is higher in the surrounding solution, then water goes inside of the cell. The movement of liquid inside or outside of the cell occurs due to the semi-permeable membrane of the cell.

So we can conclude that the specific process by which the water leaves the cell is called plasmolysis.

Learn more about cell here: brainly.com/question/13123319

#SPJ1

7 0

2 years ago

"KATZPSEF1 7.P.053.MI. MY NOTES ASK YOUR TEACHER PRACTICE ANOTHER Two black holes (the remains of exploded stars), separated by

eduard

Answer:

There are two possible solutions.

M1 = 4.68*10^30kg, M2 = 5.53*10^30kg

M1 = 5.53*10^30kg, M2 = 4.7*10^29kg

Explanation:

In order to find the mass of each black hole, you take into account the gravitational force between them and the sum of their masses.

You use the formula for the gravitational force between two masses:

$F_g=G\frac{M_1M_2}{r^2}$ (1)

G: Cavendish's constant = 6.674*10^-11 m^3/kg.s^2

M1, M2: mass of each black hole = ?

r: distance between the black holes = 10.0 AU = 10.0(1.50*10^11m) = 1.5*10^12m

Fg: gravitational force between the black holes = 7.70*10^25N

Furthermore, you take into account that the sum of the masses M1 and M2 is:

M1 + M2 = 6.00*10^30 kg (2)

You solve the equation (2) for M2.

$M_2=6.00*10^{30}-M_1$

Next, you replace the obtained expression for M2 into the equation (1) and solve for M1, as follow (for simplicity, you do not add the units):

$F_g=G\frac{M_1(6.00*10^{30}-M_1)}{r^2}\\\\\frac{r^2F_g}{G}=6.00*10^{30}M_1-M_1^2\\\\\frac{(1.5*10^{12})^2(7.70*10^{25})}{(6.674*10^{-11}}=6.00*10^{30}M_1-M_1^2\\\\2.59*10^{60}=6.00*10^{30}M_1-M_1^2\\\\M_1^2-6.00*10^{30}M_1+2.59*10^{60}=0$

Then, you have obtained a quadratic polynomial. You solve it with the quadratic formula:

$M_1=\frac{-(-6.00*10^{30})\pm \sqrt{(-6.00*10^{30})^2-4(1)(2.59*10^{60}))}}{2(1)}\\\\M_1=\frac{6.00*10^{30}\pm 5.06*10^{30}}{2}\\\\M_1=4.68*10^{29}\\\\M_1=5.53*10^{30}$