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Pavel [41]
3 years ago
13

Which is an example of a deletion mutation that results in a new protein? CAT - CGA - TCC mutates to CAC - TCG - ATC - C AAT - C

GG - CCA mutates to AAC - CGG - CCA AAT - CGG - CCA mutates to AAT - CGG - AAT - CCA CAT - CGA - TCC mutates to CAT - TCC
Chemistry
2 answers:
Tom [10]3 years ago
8 0

Answer:

D. the answer is CAT-CGA-TCC mutates to CAT-TC

Explanation:

lara31 [8.8K]3 years ago
3 0

Answer:

took the test but the answer is CAT-CGA-TCC mutates to CAT-TCC or in another way its last one d hope this helps

Explanation:

You might be interested in
Final Temperature in Heating Applesauce. A mixture of 454 kg of applesauce at 10°C is heated in a heat exchanger by adding 12130
jok3333 [9.3K]

Answer:

T_f = 76.46°C

Explanation:

Given data:

Mass of mixture = 454 kg

Initial temperature is 10°C

Heat added is Q = 121300 kJ

Heat capacity (Applesuace) at 32.8°C is 4.02kJ/kg K

From heat equation we have

Q = mCp(T_f -T_i)

\frac{Q}{mCp} = (T_f -T_i)

T_f = T_i + \frac{Q}{mCp}

Putting all value to get required final temperature value

T_f = \frac{121300}{454\times 4.02} + 10

T_f = 76.46°C

7 0
3 years ago
In the laboratory, hydrogen gas is usually made by the following reaction: Zn(s) + 2 HCl(aq) → H2(g) + ZnCl2(aq) How many liters
IrinaK [193]

<u>Answer:</u> The volume of hydrogen gas collected over water is 2.13 L

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}

Given mass of zinc = 5.566 g

Molar mass of zinc = 65.4 g/mol

Putting values in above equation, we get:

\text{Moles of zinc}=\frac{5.566g}{65.4g/mol}=0.0851mol

For the given chemical reaction:

Zn(s)+2HCl(aq.)\rightarrow H_2(g)+ZnCl_2(aq.)

As, HCl is present in excess. So, it is considered as an excess reagent.

Zinc is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

1 mole of zinc produces 1 mole of hydrogen gas.

So, 0.0851 moles of zinc will produce = \frac{1}[1}\times 0.0851=0.0851mol of hydrogen gas

To calculate the volume of hydrogen gas, we use ideal gas equation, which is:

PV = nRT

where,

P = Pressure of hydrogen gas = Total atmospheric pressure - vapor pressure of water = (752 - 18.65) mmHg = 733.35 mmHg

V = Volume of the hydrogen gas

n = number of moles of gas = 0.0851 moles

R = Gas constant = 62.3637\text{ L.mmHg }mol^{-1}K^{-1}

T = Temperature of hydrogen gas = 21^oC=[21+273]K=294K

Putting values in above equation, we get:

733.35mmHg\times V=0.0851mol\times 62.3637\text{ L.mmHg }mol^{-1}K^{-1}\times 294K\\\\V=\frac{0.0851\times 62.3637\times 294}{733.35}=2.13L

Hence, the volume of hydrogen gas collected over water is 2.13 L

5 0
3 years ago
Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur
Marina CMI [18]

Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of C_5H_{12} = 21.9 g

Molar mass of C_5H_{12} = 72.15 g/mole

Molar mass of H_2O = 18 g/mole

First we have to calculate the moles of C_5H_{12}.

\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles

Now we have to calculate the moles of H_2O.

The balanced chemical reaction will be,

C_5H_{12}(l)+8O_2(g)\rightarrow 5CO_2(g)+6H_2O(l)

From the balanced reaction we conclude that

As, 1 mole of C_5H_{12} react to give 6 moles of H_2O

So, 0.3035 moles of C_5H_{12} react to give 0.3035\times 6=1.821 moles of H_2O

Now we have to calculate the mass of H_2O.

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g

Therefore, the mass of water produced will be 32.78 grams.

3 0
3 years ago
What is the pOH of a solution of HNO3 that has [OH-] = 9.50 10-9 M?
bekas [8.4K]

Answer:

The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

Explanation:

Given data:

[OH⁻] = 9.50 ×10⁻⁹M

pOH = ?

Solution:

pOH = -log[OH⁻]

Now we will put the value of OH⁻ concentration.

pOH = -log[9.50 ×10⁻⁹M]

pOH = 8

Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.

6 0
3 years ago
Read 2 more answers
How much volume would a 834.01g pile of sugar have given that it has a density of 1.59g/mL?
boyakko [2]

Answer:

v = 534.5mL

m = 597.15g

Density = 9.23g/mL

Density = 9.125g/mL

Explanation:

Density = mass/ volume

For the first question

Density = 1.59g/mL

Mass = 834.01g

Volume = ?

Using the above formula we have 1.59 = 834.01/v

v = 834.01/1.59

v = 534.5mL

For the second question

Density =0.9167g/mL

Volume = 651.41mL

Mass =?

Using the above formula we have

0.9167 =m/651.41

Cross multiply

m = 0.9167 x 651.41

m = 597.15g

For the third question

Mass =803.44g

Volume=87.03mL

Density =?

Density = 803.44/87.03

= 9.23g/mL

For the fourth

Density = 56.85/6.23

= 9.125g/mL

7 0
3 years ago
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